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NSSCAS Physics (8225) study materials
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NSSCAS Physics (8225) practice questions

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Question 1Mechanics and kinematics

A car travels 150 m due east in 12 s and then 90 m due west in 8 s. What is the magnitude of its average velocity for the whole journey?

  1. 3.0 m/s Correct answer.
  2. 7.5 m/s
  3. 2.0 m/s
  4. 12.5 m/s

Explanation

Average velocity uses total displacement divided by total time. The displacement is 150 − 90 = 60 m east, and the total time is 20 s, giving 60/20 = 3.0 m/s. Distance travelled is not used for velocity.

Question 2Mechanics and kinematics

Which quantity is a vector?

  1. Speed
  2. Distance
  3. Time
  4. Displacement Correct answer.

Explanation

A vector needs both magnitude and direction. Displacement specifies how far and in which direction an object has moved, so it is a vector. Speed, distance and time are scalars with magnitude only.

Question 3Mechanics and kinematics

A stone is dropped from rest and falls freely. Taking g = 9.8 m/s² and ignoring air resistance, how far does it fall in the first 3.0 s?

  1. 29.4 m
  2. 44.1 m Correct answer.
  3. 14.7 m
  4. 88.2 m

Explanation

Using s = ut + ½at² with u = 0 gives s = ½ × 9.8 × 3.0² = 44.1 m. The initial velocity is zero because the stone is released from rest.

Question 4Mechanics and kinematics

On a velocity–time graph, what does the area between the line and the time axis represent?

  1. Average speed
  2. Acceleration
  3. Displacement Correct answer.
  4. Jerk

Explanation

The area under a velocity–time graph is the product of velocity and time, which equals displacement. Areas below the time axis count as negative displacement. The gradient of the graph, not the area, gives acceleration.

Question 5Mechanics and kinematics

A runner completes one full lap of a circular track of radius 50 m in 40 s. What is the magnitude of the average velocity?

  1. 7.85 m/s
  2. 0 m/s Correct answer.
  3. 3.93 m/s
  4. 1.25 m/s

Explanation

Average velocity equals displacement divided by time. After one complete lap the runner returns to the starting point, so the displacement is zero and the average velocity is zero, even though the average speed is not zero.

Question 6Mechanics and kinematics

Which statement describes the motion of an object whose acceleration is constant and in the opposite direction to its velocity?

  1. It speeds up in the direction of motion.
  2. It slows down while continuing in the same direction. Correct answer.
  3. It immediately reverses direction.
  4. It moves at constant speed.

Explanation

Acceleration opposite to velocity reduces the magnitude of the velocity, so the object decelerates. It keeps moving in the same direction until the velocity reaches zero. Reversal only happens after the velocity passes through zero.

Question 7Mechanics and kinematics

A trolley moves 4.0 m north and then 3.0 m east. What is the magnitude of its resultant displacement?

  1. 7.0 m
  2. 5.0 m Correct answer.
  3. 1.0 m
  4. 25 m

Explanation

The two displacements are perpendicular, so the resultant magnitude is found with Pythagoras: √(4.0² + 3.0²) = 5.0 m. Adding the magnitudes directly would ignore the directions.

Question 8Mechanics and kinematics

A body moves with uniform acceleration. Which graph of displacement against time would represent this motion?

  1. A straight line through the origin
  2. A horizontal straight line
  3. A curve whose gradient changes steadily Correct answer.
  4. A straight line with negative gradient

Explanation

For uniform acceleration the displacement is quadratic in time, so the displacement–time graph is a curve whose gradient equals the changing velocity. A straight line would indicate constant velocity, not constant acceleration.

Question 9Mechanics and kinematics

A cyclist accelerates uniformly from 4.0 m/s to 12.0 m/s in 5.0 s. What distance does the cyclist cover in this time?

  1. 60 m
  2. 16 m
  3. 40 m Correct answer.
  4. 80 m

Explanation

With uniform acceleration the average velocity is (4.0 + 12.0)/2 = 8.0 m/s. Multiplying by 5.0 s gives a distance of 40 m. Alternatively, using s = ut + ½at² with a = 1.6 m/s² gives the same result.

Question 10Mechanics and kinematics

A ball is thrown vertically upwards at 20 m/s. Taking g = 10 m/s², what is the maximum height reached?

  1. 10 m
  2. 2 m
  3. 40 m
  4. 20 m Correct answer.

Explanation

At maximum height the velocity is zero. Using v² = u² − 2gs with v = 0 gives s = u²/(2g) = 400/20 = 20 m. The acceleration is g downwards throughout the motion.

Question 11Mechanics and kinematics

A projectile is launched horizontally from a cliff at 15 m/s and lands 3.0 s later. Taking g = 9.8 m/s², what is the horizontal range?

  1. 29 m
  2. 44 m
  3. 15 m
  4. 45 m Correct answer.

Explanation

Horizontal and vertical motions are independent. The horizontal velocity stays at 15 m/s, so the range is 15 × 3.0 = 45 m. The vertical fall does not affect the horizontal distance.

Question 12Mechanics and kinematics

Two forces of 6.0 N and 8.0 N act at a point at right angles to each other. What is the magnitude of their resultant?

  1. 10.0 N Correct answer.
  2. 48 N
  3. 2.0 N
  4. 14.0 N

Explanation

Perpendicular forces combine by Pythagoras: √(6.0² + 8.0²) = √100 = 10.0 N. The resultant lies between the two forces in direction.

Question 13Mechanics and kinematics

A car decelerates uniformly from 25 m/s and stops in 50 m. What is the magnitude of its deceleration?

  1. 0.5 m/s²
  2. 12.5 m/s²
  3. 6.25 m/s² Correct answer.
  4. 2.0 m/s²

Explanation

Using v² = u² + 2as with v = 0 gives 0 = 625 + 2a(50), so a = −6.25 m/s². The magnitude of the deceleration is therefore 6.25 m/s².

Question 14Mechanics and kinematics

An object moves with constant speed in a circular path. Which statement about its acceleration is correct?

  1. It is directed away from the centre of the circle.
  2. It is zero because the speed is constant.
  3. It is directed towards the centre of the circle. Correct answer.
  4. It is directed along the tangent to the path.

Explanation

The velocity changes direction continuously, so there is a centripetal acceleration directed towards the centre of the circular path. Constant speed means only the direction of velocity changes, not its magnitude.

Question 15Mechanics and kinematics

A velocity–time graph shows a straight line from 0 m/s at t = 0 to 20 m/s at t = 4.0 s, then a horizontal line at 20 m/s from t = 4.0 s to t = 10 s. What is the total displacement?

  1. 40 m
  2. 160 m Correct answer.
  3. 120 m
  4. 200 m

Explanation

The displacement is the area under the graph. The triangular part gives ½ × 4.0 × 20 = 40 m, and the rectangle gives 20 × 6.0 = 120 m. Adding gives 160 m.

Question 16Mechanics and kinematics

A ball is thrown vertically upwards and returns to the thrower's hand. Ignoring air resistance, which statement is correct at the highest point?

  1. Velocity is zero and acceleration is g downwards. Correct answer.
  2. Velocity is g and acceleration is zero.
  3. Both velocity and acceleration are zero.
  4. Velocity is zero and acceleration is g upwards.

Explanation

At the highest point the ball momentarily stops, so its velocity is zero. Gravity still acts, so the acceleration remains g directed downwards throughout the flight.

Question 17Mechanics and kinematics

A boat can travel at 5.0 m/s in still water. It heads directly across a river in which the current flows at 3.0 m/s. What is the magnitude of the boat's resultant velocity relative to the bank?

  1. 2.0 m/s
  2. 8.0 m/s
  3. 5.8 m/s Correct answer.
  4. 4.0 m/s

Explanation

The boat's velocity and the current are perpendicular, so the resultant is √(5.0² + 3.0²) = √34 ≈ 5.8 m/s. The vectors must be combined by Pythagoras, not added arithmetically.

Question 18Mechanics and kinematics

A car starts from rest and accelerates at 2.0 m/s² for 6.0 s. It then travels at constant velocity for 4.0 s. What is the total distance travelled?

  1. 36 m
  2. 120 m
  3. 60 m
  4. 84 m Correct answer.

Explanation

During acceleration the distance is ½ × 2.0 × 6.0² = 36 m and the final velocity is 12 m/s. During the constant-velocity phase the distance is 12 × 4.0 = 48 m. The total is 84 m.

Question 19Mechanics and kinematics

A displacement–time graph is a curve whose gradient increases with time. What does this indicate about the motion?

  1. Constant velocity in the positive direction
  2. Constant velocity in the negative direction
  3. Acceleration in the direction of motion Correct answer.
  4. Deceleration while moving forward

Explanation

The gradient of a displacement–time graph is the velocity. An increasing gradient means the velocity is increasing, so the object is accelerating in the direction of motion.

Question 20Mechanics and kinematics

A ball is projected at 20 m/s at 30° above the horizontal. Taking g = 9.8 m/s² and ignoring air resistance, what is the time of flight on level ground?

  1. 3.5 s
  2. 2.0 s Correct answer.
  3. 4.0 s
  4. 1.0 s

Explanation

The vertical component of velocity is 20 sin 30° = 10 m/s. The time to the top is u_y/g ≈ 1.0 s, so the total flight time is about 2.0 s. Horizontal motion does not affect the flight time.

Question 21Mechanics and kinematics

A particle moves in a straight line with velocity v = 3t² − 12t (SI units). At what time is its acceleration momentarily zero?

  1. t = 6 s
  2. t = 2 s Correct answer.
  3. t = 0 s
  4. t = 4 s

Explanation

Acceleration is the derivative of velocity: a = dv/dt = 6t − 12. Setting this to zero gives t = 2 s. At that instant the velocity is at a minimum, not necessarily zero.

Question 22Mechanics and kinematics

Two cars, A and B, start from the same point. Car A moves at a constant 20 m/s. Car B starts from rest and accelerates uniformly at 2.0 m/s². At what time does B catch A?

  1. 40 s
  2. 5 s
  3. 20 s Correct answer.
  4. 10 s

Explanation

Set the displacements equal: 20t = ½ × 2.0 × t², giving t² = 20t, so t = 20 s (the other solution t = 0 is the start). At this time both have travelled 400 m.

Question 23Mechanics and kinematics

A ball is dropped from a tower of height 45 m at the same instant a second ball is thrown horizontally from the same height. Ignoring air resistance, which ball reaches the ground first?

  1. The horizontally thrown ball
  2. It depends on the horizontal speed
  3. They reach the ground at the same time Correct answer.
  4. The dropped ball

Explanation

Vertical and horizontal motions are independent. Both balls start with zero vertical velocity and fall the same vertical distance under gravity, so they take the same time. The horizontal motion does not change the vertical fall time.

Question 24Mechanics and kinematics

A particle's velocity–time graph crosses the time axis from positive to negative. What happens at the crossing point?

  1. The displacement becomes permanently zero.
  2. The acceleration changes sign.
  3. The particle is momentarily at rest and then reverses direction. Correct answer.
  4. The particle is at its maximum acceleration.

Explanation

Velocity is zero where the graph crosses the time axis, so the particle momentarily stops. After the crossing the velocity is negative, meaning the particle reverses direction. The acceleration is given by the gradient, which need not change sign there.

Question 25Mechanics and kinematics

A projectile is fired at speed u at angle θ to the horizontal on level ground. Which expression gives its horizontal range, ignoring air resistance?

  1. u² sin²θ / (2g)
  2. 2u sin θ / g
  3. u² sin 2θ / g Correct answer.
  4. u² sin θ / g

Explanation

The range is the horizontal velocity u cos θ multiplied by the flight time 2u sin θ / g, which simplifies to u² sin 2θ / g. The range is maximum at θ = 45°.

Question 26Mechanics and kinematics

A stone is thrown horizontally from the top of a vertical cliff and air resistance is negligible. Which statement describes the horizontal and vertical components of the stone's acceleration during the flight?

  1. Horizontal acceleration is zero and vertical acceleration is constant and downwards. Correct answer.
  2. Horizontal acceleration is zero and vertical acceleration is zero.
  3. Horizontal acceleration is constant and vertical acceleration is constant and downwards.
  4. Horizontal acceleration is constant and vertical acceleration increases with time.

Explanation

With air resistance negligible, no horizontal force acts on the stone once it has left the cliff, so its horizontal velocity stays constant and horizontal acceleration is zero. The only force is weight, which gives a constant vertical acceleration g directed downwards. The two components are independent, so the horizontal motion does not affect the vertical acceleration.

Question 27Mechanics and kinematics

A hiker walks 5.0 km due east and then 12 km due north. What are the magnitude and direction of the resultant displacement?

  1. 13 km on a bearing of 023°
  2. 17 km on a bearing of 023°
  3. 13 km on a bearing of 067° Correct answer.
  4. 17 km on a bearing of 067°

Explanation

The two legs are perpendicular, so the magnitude follows from Pythagoras: the square root of (5.0² + 12²) is 13 km. The direction is measured from north towards east, giving tan θ = 5.0/12, so θ is about 23° east of north, that is a bearing of 067°. Bearing is measured clockwise from north, so the angle must be taken from the north leg.

Question 28Mechanics and kinematics

A particle moves along a straight line with velocity v = 4t³ − 6t² (SI units). What is the magnitude of its acceleration at t = 2.0 s?

  1. 24 m/s²
  2. 48 m/s²
  3. 20 m/s²
  4. 36 m/s² Correct answer.

Explanation

Acceleration is the rate of change of velocity, so differentiate: a = dv/dt = 12t² − 12t. Substituting t = 2.0 s gives 12(4) − 12(2) = 48 − 24 = 24 m/s², and the magnitude is 24 m/s². The value 36 m/s² comes from substituting into the velocity expression rather than the acceleration expression.

Question 29Mechanics and kinematics

A ball is rolled up a smooth inclined plane with an initial speed of 6.0 m/s. It experiences a constant acceleration of 2.0 m/s² directed down the slope. How far up the slope does the ball travel before it momentarily comes to rest?

  1. 9.0 m Correct answer.
  2. 18 m
  3. 6.0 m
  4. 3.0 m

Explanation

Take up the slope as positive, so u = 6.0 m/s, v = 0 and a = −2.0 m/s². Using v² = u² + 2as gives 0 = 36 − 4.0s, so s = 9.0 m. The same result follows from the time to stop, 3.0 s, multiplied by the average speed of 3.0 m/s.

Question 30Mechanics and kinematics

A particle moves in a straight line and its velocity–time graph is a straight line that passes through the origin with a positive gradient. A second particle moves with a velocity–time graph that is a horizontal line at the same constant value as the first particle's final velocity. Which statement correctly compares the two motions over the same time interval starting from rest?

  1. Both particles have the same displacement because their final velocities are equal.
  2. The first particle covers half the displacement of the second particle over the same interval. Correct answer.
  3. The second particle has the greater displacement because its velocity is constant throughout.
  4. The first particle has the greater displacement because its velocity is always increasing.

Explanation

For the first particle the velocity rises uniformly from zero to v, so its displacement is the triangular area, ½vt. The second particle moves at the constant velocity v for the whole time, so its displacement is the rectangular area vt. The first particle therefore covers exactly half the displacement of the second.

Question 31Forces, momentum and Newtonian mechanics

A body moves in a straight line at constant velocity. Which statement about the forces acting on it must be true?

  1. A single constant forward force acts on it.
  2. The resultant force acts perpendicular to the velocity.
  3. The forward force is larger than the backward force.
  4. The resultant force on the body is zero. Correct answer.

Explanation

Newton's first law states that a body continues at constant velocity only when the resultant force is zero. A constant non-zero resultant force would produce acceleration, changing the speed or direction. Balanced individual forces may still exist, but their vector sum must vanish.

Question 32Forces, momentum and Newtonian mechanics

A resultant force of 24 N acts on a mass of 6.0 kg. What is the acceleration of the mass?

  1. 4.0 m s^-2 Correct answer.
  2. 0.25 m s^-2
  3. 18 m s^-2
  4. 144 m s^-2

Explanation

Newton's second law gives a = F/m. Dividing 24 N by 6.0 kg yields 4.0 m s^-2. The acceleration is in the same direction as the resultant force.

Question 33Forces, momentum and Newtonian mechanics

Which pair of quantities are both vectors?

  1. Work and momentum
  2. Mass and weight
  3. Momentum and weight Correct answer.
  4. Speed and velocity

Explanation

Momentum (mass times velocity) and weight (mass times gravitational field strength) both have direction as well as magnitude, so both are vectors. Mass, speed, work and energy are scalars. Recognising this distinction matters when adding forces and momenta vectorially.

Question 34Forces, momentum and Newtonian mechanics

A stationary ball of mass 0.15 kg is struck and leaves with speed 12 m s^-1. The contact lasts 0.040 s. What is the average force exerted on the ball?

  1. 1.8 N
  2. 0.072 N
  3. 45 N Correct answer.
  4. 0.45 N

Explanation

The change in momentum is 0.15 x 12 = 1.8 kg m s^-1. Force equals rate of change of momentum, so F = 1.8 / 0.040 = 45 N. This is the impulse-momentum form of Newton's second law.

Question 35Forces, momentum and Newtonian mechanics

Two forces of 6.0 N and 8.0 N act at a point at right angles to each other. What is the magnitude of the resultant force?

  1. 10 N Correct answer.
  2. 48 N
  3. 14 N
  4. 2.0 N

Explanation

Perpendicular forces add as the hypotenuse of a right-angled triangle, so R = sqrt(6.0^2 + 8.0^2) = sqrt(100) = 10 N. Simply adding magnitudes would only be valid if the forces were parallel.

Question 36Forces, momentum and Newtonian mechanics

A skydiver falls at terminal velocity. Which statement describes the forces on the skydiver?

  1. Drag exceeds weight, so the skydiver decelerates.
  2. Weight exceeds drag, so the skydiver accelerates downward.
  3. Weight and drag are equal in magnitude and opposite in direction. Correct answer.
  4. No forces act on the skydiver at terminal velocity.

Explanation

At terminal velocity the acceleration is zero, so the resultant force must be zero. The downward weight is exactly balanced by the upward drag force. This is why the speed stays constant thereafter.

Question 37Forces, momentum and Newtonian mechanics

A car of mass 1200 kg travelling at 20 m s^-1 brakes to rest in 4.0 s. What is the magnitude of the average braking force?

  1. 4800 N
  2. 240 N
  3. 6000 N Correct answer.
  4. 96 000 N

Explanation

The deceleration is 20/4.0 = 5.0 m s^-2. Applying F = ma gives 1200 x 5.0 = 6000 N. Alternatively, the momentum change 24 000 kg m s^-1 divided by 4.0 s gives the same force.

Question 38Forces, momentum and Newtonian mechanics

A 3.0 kg trolley moving at 2.0 m s^-1 collides with a stationary 1.0 kg trolley and they move off together. What is their common speed?

  1. 0.67 m s^-1
  2. 6.0 m s^-1
  3. 2.0 m s^-1
  4. 1.5 m s^-1 Correct answer.

Explanation

Momentum is conserved: 3.0 x 2.0 = 6.0 kg m s^-1 before, and the combined mass is 4.0 kg. Dividing gives 6.0/4.0 = 1.5 m s^-1. Kinetic energy is not conserved in a perfectly inelastic collision.

Question 39Forces, momentum and Newtonian mechanics

A force of 50 N acts on a body and moves it 3.0 m in the direction of the force. What is the work done by the force?

  1. 17 J
  2. 53 J
  3. 150 J Correct answer.
  4. 0 J

Explanation

Work done equals force times displacement in the direction of the force, so W = 50 x 3.0 = 150 J. The joule is the unit of work and energy. If the force were perpendicular to the displacement, the work would be zero.

Question 40Forces, momentum and Newtonian mechanics

A box is pulled along a rough horizontal floor at constant speed by a horizontal force. Which statement about the forces is correct?

  1. The pulling force equals the friction force in magnitude. Correct answer.
  2. The pulling force is greater than the friction force.
  3. The resultant force equals the pulling force.
  4. The normal contact force equals the pulling force.

Explanation

Constant speed means zero acceleration and therefore zero resultant force. The horizontal pulling force must be exactly balanced by kinetic friction. The normal contact force instead balances the weight.

Question 41Forces, momentum and Newtonian mechanics

A projectile is launched horizontally from a cliff and air resistance is negligible. Which statement about its motion is correct?

  1. The vertical velocity stays constant while the horizontal velocity increases.
  2. The horizontal velocity decreases while the vertical velocity increases.
  3. Both horizontal and vertical velocities stay constant.
  4. The horizontal velocity stays constant while the vertical velocity increases. Correct answer.

Explanation

With no air resistance there is no horizontal force, so the horizontal velocity remains constant. The only force is weight, giving a constant downward acceleration that increases the vertical velocity. The two components are independent.

Question 42Forces, momentum and Newtonian mechanics

A force of 12 N acts at 60 degrees to the horizontal. What is the horizontal component of the force?

  1. 6.0 N Correct answer.
  2. 12 N
  3. 10.4 N
  4. 0 N

Explanation

The horizontal component is F cos(theta) = 12 x cos 60 = 12 x 0.5 = 6.0 N. The vertical component would be F sin(theta), also 10.4 N in this case. Resolving forces is essential for equilibrium problems on inclines.

Question 43Forces, momentum and Newtonian mechanics

A ball of mass 0.20 kg hits a wall horizontally at 8.0 m s^-1 and rebounds at 6.0 m s^-1. What is the magnitude of the change in momentum of the ball?

  1. 1.6 kg m s^-1
  2. 1.2 kg m s^-1
  3. 0.40 kg m s^-1
  4. 2.8 kg m s^-1 Correct answer.

Explanation

Taking the initial direction as positive, momentum changes from +1.6 to -1.2 kg m s^-1. The change is -1.2 - 1.6 = -2.8 kg m s^-1, so the magnitude is 2.8 kg m s^-1. Reversal of direction means the speeds add in the change calculation.

Question 44Forces, momentum and Newtonian mechanics

A block slides down a smooth incline of angle 30 degrees. What is its acceleration along the incline?

  1. 8.5 m s^-2
  2. 4.9 m s^-2 Correct answer.
  3. 0 m s^-2
  4. 9.8 m s^-2

Explanation

The component of weight along a smooth incline is mg sin(theta), so the acceleration is g sin 30 = 9.8 x 0.5 = 4.9 m s^-2. The normal contact force balances the perpendicular component of weight.

Question 45Forces, momentum and Newtonian mechanics

A person of mass 70 kg stands in a lift that accelerates upward at 2.0 m s^-2. Taking g as 9.8 m s^-2, what is the normal contact force on the person?

  1. 826 N Correct answer.
  2. 140 N
  3. 686 N
  4. 546 N

Explanation

The resultant upward force is N - mg = ma, so N = m(g + a) = 70 x 11.8 = 826 N. The contact force exceeds the weight because it must both support the weight and provide the upward acceleration.

Question 46Forces, momentum and Newtonian mechanics

Two skaters push apart from rest on frictionless ice. Skater A has mass 40 kg and moves at 3.0 m s^-1. Skater B has mass 60 kg. What is the speed of skater B?

  1. 2.0 m s^-1 Correct answer.
  2. 4.5 m s^-1
  3. 3.0 m s^-1
  4. 1.5 m s^-1

Explanation

The total momentum before is zero, so it must remain zero afterward. Thus 40 x 3.0 = 60 x v, giving v = 120/60 = 2.0 m s^-1 in the opposite direction. The forces the skaters exert on each other are equal and opposite.

Question 47Forces, momentum and Newtonian mechanics

A vehicle of mass 800 kg moving at 15 m s^-1 collides with a stationary vehicle of mass 1200 kg and they lock together. What is their common velocity?

  1. 9.0 m s^-1
  2. 6.0 m s^-1 Correct answer.
  3. 15 m s^-1
  4. 10 m s^-1

Explanation

Momentum before is 800 x 15 = 12 000 kg m s^-1. The combined mass is 2000 kg, so v = 12 000/2000 = 6.0 m s^-1. The collision is inelastic, so kinetic energy decreases.

Question 48Forces, momentum and Newtonian mechanics

A crate is dragged at constant speed across a rough floor by a rope inclined at 30 degrees above the horizontal. Compared with pulling horizontally, why is the friction force smaller?

  1. The horizontal component of tension is larger than when pulling horizontally.
  2. The tension is larger, which reduces the coefficient of friction.
  3. The vertical component of tension reduces the normal contact force. Correct answer.
  4. Friction depends only on the speed of the crate.

Explanation

Friction is proportional to the normal contact force, which equals mg minus the upward component of tension. An inclined pull reduces the normal force and therefore the friction. The coefficient of friction is a property of the surfaces and does not change.

Question 49Forces, momentum and Newtonian mechanics

A force acts on a body and the body's momentum changes by 30 kg m s^-1 in 5.0 s. What is the average force?

  1. 150 N
  2. 25 N
  3. 0.17 N
  4. 6.0 N Correct answer.

Explanation

By the impulse-momentum theorem, average force equals change in momentum divided by time. Dividing 30 by 5.0 gives 6.0 N. This follows directly from F = dp/dt.

Question 50Forces, momentum and Newtonian mechanics

A body moves in a circle at constant speed. Which statement about its acceleration is correct?

  1. The acceleration is directed away from the centre of the circle.
  2. The acceleration is directed along the tangent to the circle.
  3. The acceleration is zero because the speed is constant.
  4. The acceleration is directed toward the centre of the circle. Correct answer.

Explanation

Even at constant speed, the velocity direction changes continuously, so there is a resultant force and an acceleration. This centripetal acceleration points toward the centre of the circle. It is produced by a force such as tension or gravity.