5.2. Gravitational potential energy and kinetic energy
A subsection of Physics, 9702, through 5. Work, energy and power
Listing 10 of 93 questions
An archer releases an arrow towards a target at a velocity of 65.0 m s–1 at an angle of 4.30° above the horizontal, as shown in . target centre of target ground arrow, speed 65.0 m s–1 70.0 m 1.66 m 4.30° archer (not to scale) When released, the tip of the arrow is a horizontal distance of 70.0 m from the target and 1.66 m above the horizontal ground. The arrow hits the centre of the target. Assume that air resistance is negligible and that all the mass of the arrow is at its tip. Show that the time taken for the arrow to reach the target is 1.08 s. Calculate the height of the centre of the target above the ground. height above ground = m By considering energy changes, state and explain how the final kinetic energy of the arrow as it hits the target compares with its initial kinetic energy immediately after release. A numerical calculation is not required.
9702_s22_qp_23
THEORY
2022
Paper 2, Variant 3
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State what is meant by the centre of gravity of an object. Two blocks are on a horizontal beam that is pivoted at its centre of gravity, as shown in . 54 N 2.4 N 0.35 m 0.95 m 0.45 m horizontal string pivot support beam ground 30° T (not to scale) A large block of weight 54 N is a distance of 0.45 m from the pivot. A small block of weight 2.4 N is a distance of 0.95 m from the pivot and a distance of 0.35 m from the right‑hand end of the beam. The right‑hand end of the beam is connected to the ground by a string that is at an angle of 30° to the horizontal. The beam is in equilibrium. By taking moments about the pivot, calculate the tension T in the string. T = N The string is cut so that the beam is no longer in equilibrium. Calculate the magnitude of the resultant moment about the pivot acting on the beam immediately after the string is cut. resultant moment = N m The beam in rotates when the string is cut and the small block of weight 2.4 N is projected through the air. shows the last part of the path of the block before it hits the ground at point Y. 1.8 m horizontal ground path of block X Y (not to scale) At point X on the path, the block has a speed of 3.4 m s–1 and is at a height of 1.8 m above the horizontal ground. Air resistance is negligible. Calculate the decrease in the gravitational potential energy of the block for its movement from X to Y. decrease in gravitational potential energy = J Use your answer to and conservation of energy to determine the kinetic energy of the block at Y. kinetic energy = J State the variation, if any, in the direction of the acceleration of the block as it moves from X to Y. The block passes point X at time tX and arrives at point Y at time tY. On , sketch a graph to show the variation of the magnitude of the horizontal component of the velocity of the block with time from tX to tY. Numerical values are not required. time horizontal component of velocity tX tY
9702_s23_qp_22
THEORY
2023
Paper 2, Variant 2
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Define gravitational potential at a point. The Moon may be considered to be an isolated sphere of radius 1.74 × 103 km with its mass of 7.35 × 1022 kg concentrated at its centre. A rock of mass 4.50 kg is situated on the surface of the Moon. Show that the change in gravitational potential energy of the rock in moving it from the Moon’s surface to infinity is 1.27 × 107 J. The escape speed of the rock is the minimum speed that the rock must be given when it is on the Moon’s surface so that it can escape to infinity. Use the answer in to determine the escape speed. Explain your working. speed = m s–1 The Moon in is assumed to be isolated in space. The Moon does, in fact, orbit the Earth. State and explain whether the minimum speed for the rock to reach the Earth from the surface of the Moon is different from the escape speed calculated in .
9702_w13_qp_41
THEORY
2013
Paper 4, Variant 1
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Define gravitational potential at a point. The Moon may be considered to be an isolated sphere of radius 1.74 × 103 km with its mass of 7.35 × 1022 kg concentrated at its centre. A rock of mass 4.50 kg is situated on the surface of the Moon. Show that the change in gravitational potential energy of the rock in moving it from the Moon’s surface to infinity is 1.27 × 107 J. The escape speed of the rock is the minimum speed that the rock must be given when it is on the Moon’s surface so that it can escape to infinity. Use the answer in to determine the escape speed. Explain your working. speed = m s–1 The Moon in is assumed to be isolated in space. The Moon does, in fact, orbit the Earth. State and explain whether the minimum speed for the rock to reach the Earth from the surface of the Moon is different from the escape speed calculated in .
9702_w13_qp_42
THEORY
2013
Paper 4, Variant 2
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A motor is used to move bricks vertically upwards, as shown in . motor container bricks The bricks start from rest and accelerate for 2.0 s. The bricks then travel at a constant speed of 0.64 m s−1 for 25 s. Finally the bricks are brought to rest in a further 3.0 s. The total mass of the bricks is 25 kg. Determine the change in kinetic energy of the bricks in the first 2.0 s, change in kinetic energy = J in the next 25 s, change in kinetic energy = J in the final 3.0 s. change in kinetic energy = J The bricks are in a container. The weight of the container and bricks is 350 N. Calculate, for the lifting of the bricks and container when travelling at constant speed, the gain in potential energy, energy gain = J the power required. power = W
9702_w14_qp_21
THEORY
2014
Paper 2, Variant 1
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A golfer strikes a ball so that it leaves horizontal ground with a velocity of 6.0 m s–1 at an angle θ to the horizontal, as illustrated in . ground 6.0 m s–1 4.8 m s–1 vX vY θ ball (not to scale) The magnitude of the initial vertical component vY of the velocity is 4.8 m s–1. Assume that air resistance is negligible. Show that the magnitude of the initial horizontal component vX of the velocity is 3.6 m s–1. The ball leaves the ground at time t = 0 and reaches its maximum height at t = 0.49 s. On , sketch separate lines to show the variation with time t, until the ball returns to the ground, of the vertical component vY of the velocity (label this line Y), the horizontal component vX of the velocity (label this line . –5.0 –4.0 –3.0 –2.0 –1.0 1.0 2.0 3.0 velocity / m s–1 t / s 4.0 5.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Calculate the maximum height reached by the ball. maximum height = m For the movement of the ball from the ground to its maximum height, determine the ratio kinetic energy at maximum height change in gravitational potential energy . ratio = In practice, significant air resistance acts on the ball. Explain why the actual time taken for the ball to reach maximum height is less than the time calculated when air resistance is assumed to be negligible.
9702_w18_qp_22
THEORY
2018
Paper 2, Variant 2
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Questions Discovered
93