Drayton Manor High School Momentum, Impulse, Force-time Graphs Old Exam Questions Q1. Which one of the following is a possible unit of impulse? A Ns 1 B kg ms 1 C kg ms 2 D sn 1 (Total 1 mark) Q2. Which one of the following statements is correct? The force acting on an object is equivalent to A B C D its change of momentum. the impulse it receives per second. the energy it gains per second. its acceleration per metre. Q3. A ball of mass 2.0 kg, initially at rest, is acted on by a force F which varies with time t as shown by the graph. (Total 1 mark) What is the velocity of the ball after 8.0 s? A 20 ms 1 B 40 ms 1 C 80 ms 1 D 160 ms 1 (Total 1 mark) Page 1
Q4. The graph shows how the force acting on a body changes with time. Drayton Manor High School The body has a mass of 0.25 kg and is initially at rest. What is the speed of the body after 40 s assuming no other forces are acting? A 200 ms 1 B 400 ms 1 C 800 ms 1 D 1600 ms 1 (Total 1 mark) Q5. The diagram shows the graph of force on a car against time when the car of mass 500 kg crashed into a wall without rebounding. Which one of the following statements is correct? A B C The area under the graph is equal to the initial momentum of the car Momentum is not conserved in the collision Kinetic energy is conserved in the collision D The average force exerted on the car is 10 10 4 N (Total 1 mark) Page 2
Q6. The graph shows how the resultant force, F, acting on a body varies with time, t. Drayton Manor High School What is the change in momentum of the body over the 5 s period? A B C D 2N s 8N s 10N s 12N s (Total 1 mark) Q7. A body is accelerated from rest by a constant force. Which one of the following graphs best represents the variation of the body s momentum p with time t? (Total 1 mark) Page 3
Drayton Manor High School Q8. A toy locomotive of mass 0.50 kg is initially at rest on a horizontal track. The locomotive is powered by a twisted rubber band which, as it unwinds, exerts a force which varies with time as shown in the table. time/s 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 force/n 0.20 0.18 0.15 0.12 0.10 0.08 0.05 0.02 0.00 (a) (i) On the grid below plot a graph of force against time for the rubber band power source. Page 4
(ii) State what is given by the area between the graph and the time axis. Drayton Manor High School (4) (b) The rubber band is wound up and released to power the locomotive. Use your graph to show that the speed of the locomotive 8.0 s after the twisted rubber band is released is 1.6 m s 1. Ignore the effects of air resistance and energy losses due to friction. (c) 8.0 s after release the locomotive collides with and couples to a toy truck, initially at rest, which has a mass of 1.50 kg. (i) Calculate the speed of the coupled locomotive and truck after collision. (ii) Calculate the combined kinetic energy of the locomotive and truck immediately after collision. (iii) Show, with the aid of a calculation, whether or not the collision is elastic. (5) (Total 11 marks) Q9. The graph shows the variation in the horizontal force acting on a tennis ball with time whilst the ball is being served. Page 5
Drayton Manor High School (a) (i) Use the graph to show that the magnitude of the impulse that acts on the tennis ball is about 1.3 N s. (ii) The mass of the tennis ball is 0.057 kg. Show that the impulse in part (a)(i) gives the ball a speed of about 20 m s 1 horizontally as the ball leaves the racquet. Assume that the ball had no horizontal speed before the impulse was applied. (b) During flight the ball accelerates due to gravity. When it reaches the ground the vertical component of the velocity is 6.1 m s 1. Calculate the speed and the angle between the direction of travel of the ball and the horizontal as it reaches the ground. Assume that air resistance is negligible. speed... m s 1 angle... degree (Total 7 marks) Q10. (a) State, in words, how the force acting on a body is related to the change in momentum of the body....... (1) (b) A football of mass 0.42 kg is moving horizontally at 10 m s 1 towards a footballer s boot, which then kicks it. The figure below shows how the force between the boot and the ball varies with time while they are in contact. Page 6
Drayton Manor High School (i) What is the significance of the area enclosed by the line on a force time graph and the time axis when a force acts on a body for a short time?... (1) (ii) Estimate the impulse that acts on the ball, stating an appropriate unit. answer =... (4) (iii) Calculate the speed of the ball after it has been kicked, assuming that it returns along the same horizontal line it followed when approaching the boot. Express your answer to an appropriate number of significant figures. answer =...m s 1 (4) (c) Discuss the consequences if the ball had approached the boot at a higher speed but still received the same impulse.... Page 7
.................. Drayton Manor High School... (Total 13 marks) Q11. (a) A car, of mass 970 kg, is travelling at 15 m s 1 along a level road when its driver performs an emergency stop. The car s braking system applies a constant braking force of 6.1 10 3 N to the car. Assume that the braking force is the resultant force acting on the car. (i) Calculate the change of momentum of the car during the emergency stop. change of momentum... kg m s 1 (1) (ii) Calculate the distance the car moves in coming to a halt during the emergency stop. distance... m (4) (b) The car is now loaded with passengers and luggage and again travels at 15 m s 1. State and explain how this affects the braking distance of the car. Assume that the car experiences the same braking force as in part (a). Page 8
Drayton Manor High School (Total 8 marks) Q12. (a) State the principle of conservation of momentum. (b) The diagram below shows a sketch drawn by an accident investigator following a head-on collision between two vehicles. From the skid marks and debris on the road the investigator knows that the collision took place at the point marked X. The vehicles locked together on impact and vehicle A was pushed backwards a distance of 8.4 m. For the road conditions and vehicle masses the average frictional force between the road and the vehicles immediately after the collision was known to be 7500 N. (i) Calculate the work done against friction in bringing the vehicles to rest. (ii) Determine the speed of the interlocked vehicles immediately after impact. Page 9
Drayton Manor High School (iii) Vehicle A was known to be moving at 12.5 m s 1 just before the impact. Calculate the speed of vehicle B just before impact. (iv) The drivers of both vehicles have the same mass. State and explain which driver is likely to experience the higher force during the impact. (4) (Total 13 marks) Page 10
Drayton Manor High School Q13. (a) Starting with the relationship between impulse and the change in momentum, show clearly that the unit, N, is equivalent to kg m s 2. (b) A rocket uses a liquid propellant in order to move. Explain how the ejection of the waste gases in one direction makes the rocket move in the opposite direction. (c) A rocket ejects 1.5 10 4 kg of waste gas per second. The gas is ejected with a speed of 2.4 km s 1 relative to the rocket. Show that the average thrust on the rocket is about 40 MN. (Total 7 marks) Q14. (a) Explain what is meant by the principle of conservation of momentum. Page 11
Drayton Manor High School (b) A hose pipe is used to water a garden. The supply delivers water at a rate of 0.31 kg s 1 to the nozzle which has a cross-sectional area of 7.3 10 5 m 2. (i) Show that water leaves the nozzle at a speed of about 4 m s 1. density of water = 1000 kg m 3 (ii) Before it leaves the hose, the water has a speed of 0.68 m s 1. Calculate the force on the hose. (iii) The water from the hose is sprayed onto a brick wall the base of which is firmly embedded in the ground. Explain why there is no overall effect on the rotation of the Earth. (Total 9 marks) Page 12
Drayton Manor High School Q15. The figure below shows a neutron of mass 1.7 10 27 kg about to collide inelastically with a stationary uranium nucleus of mass 4.0 10 25 kg. During the collision, the neutron will be absorbed by the uranium nucleus. (a) Calculate the velocity of the uranium nucleus immediately after the neutron has been absorbed. (b) Collisions between neutrons and uranium nuclei can also be elastic. State, and explain briefly, how the speed of the uranium nucleus after impact would be different in the case of an elastic collision. Do not perform any further calculations. (c) Using the data at the beginning of the question, calculate the kinetic energy of the neutron before it collides with the uranium nucleus. (Total 9 marks) Page 13
Drayton Manor High School ANSWERS M1. B M2. B M3. A M4. C M5. A M6. C M7. B M8.(a) (i) points plotted correctly (1) (1) (deduct one for each incorrect) sensible scales chosen (1) line of best fit (1) [1] [1] [1] [1] [1] [1] [1] (ii) change in momentum [or impulse] (1) (accept 0.8) (b) area under graph = 0.80±0.05 (1) (kgms 1 ) max 4 υ = (1) = 1.6 ms 1 alternative: state average force = 0.10(N) (1) leading to correct derivation of 1.6ms 1 (1) 2 (c) (i) mυ = 0 [or statement] (1) υ = 0.40ms -1 (1) (ii) kinetic energy = 0.16 J (1) (iii) initial kinetic energy = 0.64 (J) (1) kinetic energy lost so inelastic (1) 5 [11] M9. (a) (i) clear working: attempts to evaluate area under line (12 13 squares 0.1 N s) or ½ 106(105) 24 10 3 seen 1.260 1.272 (N s) 1.2 1.3 for square counting (ii) area under graph/impulse = mv Page 14
22.3 (22.8) (m s 1 ) Drayton Manor High School (b) use of Pythagoras or tan 1 (6.1/22.3) or or tan 1 (6.1/20) 22.8 23.8(m s 1 ) 20.9 or 21(m s 1 ) 14.9 15.5 ( ) (15 16 ( )) 17/18( ) 3 [7] M10. (a) force = rate of change of momentum (1) (b) (i) area under graph represents impulse or change in momentum (1) 1 1 (ii) suitable method to estimate area under graph (1)(1) [eg counting squares: 20 to 23 squares (1) each of area 25 10 3 20 = 0.5 (N s) (1) or approximate triangle etc (1) ½ 250 10 3 90 (1)] gives impulse = 11 ± 1 (1) N s (or kg m s 1 ) (1) 4 (iii) use of impulse = (mv) (1) p = mv ( mu) = m(v + u) or 11 = 0.42 (v + 10) (1) giving 0.42 v = 6.8 and v = 16 (m s 1 ) (impulse = 12 gives 19 m s 1 ) (1) answer to 2 sf only (1) 4 (c) final speed would be lower (1) any two of the following points (1)(1) initial momentum would be greater [or greater u must be reversed] change in momentum [or velocity] is the same [or larger F acts for shorter t] initial and final momenta are (usually) in opposite directions initial and final momenta may be in same direction if initial speed is sufficiently high [alternatively] final speed = initial speed (1) Page 15
gives final speed v = (26 ± 3) initial speed u (1) Drayton Manor High School consequence is v is in opposite direction to u when u < 26 v is in same direction as u when u > 26 v is zero (ball stationary) when u = 26 any one of these bullet points (1) 3 [13] M11. (a) (i) 1.5 10 4 /1.46 10 4 (14550) (kg m s 1 ) 1 (ii) correct substitution into t = (mv mu)/f or (t =) 14550 6.1 10 3 seen (condone power of 10 error) [ecf from part ai] C1 t = 2.39 s [ecf from part ai] C1 correct substitution into s = (u + v)t/2 or (s =) 15 2.39 2 seen (condone incorrect value for a calculated t in substitution) C1 s = 17.9 to 18 m [ecf from part ai] (b) (braking distance) increases/ longer to stop 4 M0 greater mass more momentum/more time with rate of change of momentum equation same velocity change over longer time means greater distance /appropriate equation of motion with s as subject and longer time when using s = vt must identify v as average or (braking distance) increases/ longer to stop M0 greater mass Page 16
more ke (to convert)/more work to be done Drayton Manor High School more work done by (same) force means greater distance/ appropriate equation with s as subject or (braking distance) increases/ longer to stop M0 greater mass smaller acceleration smaller acceleration for (same) change in velocity means greater distance/appropriate equation of motion with s as subject 3 [8] M12. (a) total momentum before a collision = total momentum after a collision or total momentum of a system is constant or Σmv = 0, where mv is the momentum no external forces acting on the system/ isolated system 2 (b) (i) work done = F s C1 63 000 J 2 (ii) KE = ½ mv 2 or F =ma and v 2 = u 2 + 2as C1 combined speed v = 4.6 (4.58) m s 1 2 (iii) reasonable attempt at a momentum conservation equation (2 terms before and one term after any signs) C1 (+ or -) 3600 v + (2400 12.5) = (6000 4.58) (e.c.f) C1 16 m s 1 (cao ignoring sign) 3 (iv) driver A is likely to experience the greater force Page 17
force = rate of change of momentum ( mv/t) or F = ma Drayton Manor High School time for deceleration on impact is (approximately) the same change in velocity of driver B = 11.4 m s 1 (ecf from (ii) and (iii)) and change in velocity of driver A = 17.1 m s 1 (ecf from (ii) and (iii)) or mv or v of A > mv or v of B 4 [13] M13. (a) F = or Ft = mv mu etc. M1 substitute units 2 (b) conservation of momentum mentioned ejected gas has momentum or velocity in one direction rocket must have equal momentum in the opposite direction 3 or force = rate of change of momentum () ejected gas has momentum or velocity in one direction () rocket must have equal and opposite force () (c) equation seen (F = m/t v but not F= ma) substitution into any sensible equation leading to 3.6 10 7 (N) 2 [7] M14. (a) total momentum of system constant/total momentum before = total momentum after isolated system/no external force Page 18
Drayton Manor High School 2 (b) (i) clear explanation of method correct numerical working leading to 4.25 m s 1 2 (ii) F = 0.31 a speed C1 use of speed difference [4.25 0.68] C1 = 1.11 N [ecf] 3 (iii) states that two momenta/forces related to hose and wall are equal in size/appreciates reaction force transmitted by hose to Earth and in opposite direction 2 [9] M15.(a) conservation of momentum equation or statement quoted or used even with incorrect data 1.4 10 7 1.7 10 27 = 401.7 10 27 v C1 C1 5.9 10 4 m s 1 (b) neutron will rebound / have velocity / momentum to the left momentum transferred to the uranium will be greater Page 19
velocity of uranium will be greater Drayton Manor High School (no loss of kinetic energy argument gets the final mark only) (c) KE = ½mv 2 seen or used C1 0.5 1.7 10 27 (1.4 10 7 ) 2 C1 1.7 10 13 J (1.67 10 13 J) [9] Page 20