JPJC 2026 Collisions Tutorial Solutions

1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS COLLISIONS TUTORIAL SOLUTIONS Self-Check Questions S1 Impulse of a force is the product of the force and the time interval for which the force acts on the body. It is mathematically equal to the area under the force-time graph. S2 The principle of conservation of momentum states that the total momentum of a system remains constant provided no net external force acts on it. S3 Consider a moving object A, of mass mA and velocity uA, collides with an object B, of mass mB and velocity uB, moving in the same direction as in Fig. 3.1. Fig. 3.1 Fig. 3.2 After the collision, they move off with velocities vA and vB as shown in Fig. 3.2. During the brief collision, these objects exert large forces on one another. By Newton’s third law, the force FBA exerted by B on A is equal in magnitude and opposite in direction to the force FAB exerted by A on B, as shown in Fig. 3.3. Fig. 3.3   BA AB A B A A A A B B B B A A A A B B B B A A B B A A B B F F p p t t m v m u m v m u t t m v m u m v m u m u m u m v m v              Total initial momentum of bodies = Total final momentum of bodies (provided no net external force act on bodies A and B) S4 Kinetic energy is conserved in elastic collisions but not in inelastic collisions. Objects stick together in completely inelastic collisions. For all collisions, the total momentum and total energy of the system are conserved.

2 Self-Practice Questions Qn Ans Solution P1 10 N Area under F-t graph = impulse = change in momentum  1 5 2 352 10 N x x      P2 C By conservation of linear momentum, Total initial moment of system = total final momentum of the system Hence, total momentum     12 12 4 0 24 kg m s      P3 A Before release, the system of trolleys X and Y has a total momentum of zero. With no external resultant force acting on the system, the total momentum of the system remains at zero. So, after collision, both trolleys must remain stationary in order for the total momentum to be zero. Hence, the speed of Y after collision is zero. P4 C Principle of conservation of momentum 0.30(1.2) 0.60( 1.8) 0.30 0.60 P Qv v    2 2.4Q Pv v   Relative speed of approach = Relative speed of separation 1.2 ( 1.8) Q Pv v    3.0Q Pv v  Solving 12 .8 m sPv   10.2 m sQv 

3 Discussion Questions 1 After collision, let velocity of mass m be vm and velocity of mass M be vM. By conservation of linear momentum, --- (1)m Mmv Mv mu  By relative speed of approach = relative speed of separation 0 --- (2) M m M m u v v v v u      Substitute (2) into (1) and solve for vM. 2 M muv M m  Substitute the above into (2) to solve for vm. 2 m muv u M m  Solve for v: 2 2 M m mu muv v v u u M m M m           Therefore, v is directly proportional to u, hence, sketch a straight-line graph that cuts the origin where gradient is 1. 2 (a) (i) (ii) By Newton’s third law of motion, the force that A exerts on B is equal in magnitude but opposite in direction to the force B exerts on A. (iii) The area below the F-t graph gives the change in momentum experienced by body. The change in momentum of body B is positive while the change in momentum of body A is negative (graph is below t-axis), and they have the same magnitude. Hence, the overall change in momentum experienced by the collision system of bodies A and B is zero. 0 force time force that A exerts on B force that B exerts on A u v 0

4 Hence, this is consistent with principle of conservation of momentum where there is no change in momentum of the system as a whole, provided no external resultant force acts on the system. (b) By Impulse-Momentum Theory, Change in momentum = Impulse = Ft Change in velocity of car = 1(72000)(0.25) 15 ms1200car Ft m   Change in velocity of truck = 1(72000)(0.25) 1.5 ms12000truck Ft m     Note that the velocity of the truck decreases. (c) The force experienced by the vehicles is not constant in reality. The force varies as both the truck and car undergo deformation during a collision. (d) The seat belt and air bag provide a backward force to stop the passenger from continuing to move forward due to his inertia. This backward force is provided over a prolonged period of time. Applying Newton’s second law, this also means that the rate of change of momentum of the passenger is lower. The resultant force acting on the passenger is thus reduced, lowering the risk of injury. (e) Any three factors: - surface composition of the road - driving speed - tyre pressure - tyre quality and erosion - tyre maintenance - road visibility - road structure (e.g. curves, widths, slopes)

5 3 (a) (i) Like charges repel. (ii) There are no external forces acting on the system, hence total momentum of the system is conserved. However if both were to stop at the same time, it means the total momentum at that instant is zero, which is impossible since the initial momentum of the system is 3mv – 2mv = mv which is not zero. (b) By principle of conservation of momentum and taking right +ve: f f3 2 ( ) 3 2mv m v mv mv    f 0.2v v (c) (i) (ii) Let final velocity of A be vA and that of B be vB. Taking right as positive, By principle of conservation of momentum: A B3 2 ( ) 3 2 --- (1)mv m v mv mv    Since interaction is elastic, thus relative speed of approach = relative speed of separation B A2 --- (2)v v v  Solving both equations yields vA = – 0.6v and vB = 1.4v Thus speed of A = 0.6 v and speed of B = 1.4v Note: The negative sign of vA shows that A actually changes direction. Similarly, the fact that vB is positive shows that B has also changed direction. A and B repel each other and move away in opposite directions. 4 (a) (i) Knowing that their collision is elastic tells us that the total kinetic energy of the molecules is conserved in this collision. Moreover the relative speed of approach of the molecules equals their relative speed of separation. (ii) A head-on collision is one where the molecules remain moving along the same straight line joining their centres of mass before and after collision. (b) (i) Since collision is elastic: (taking to the right as positive) relative velocity of separation of the two molecules = relative velocity of approach relative velocity of approach = 3 1 1 2 1.88 10 ( 405) 2285 m su u       velocity time not to scale A B tC tA tB

6 Hence relative velocity of separation of the two molecules, 1 2 1 2285 m sv v   ---- (1) (ii) let the final velocity of the hydrogen molecule be v1 and that of the oxygen molecule be v2, taking right to be positive: 3 1 22.00 (1.88 10 ) 32.0 ( 405) 2.00 ( ) 32.0 ( )u u u v u v     ---- (2) Solving equations (1) and (2) yields v1 = – 2420 m s–1 , v2 = – 136 m s–1 Hydrogen molecule will move to the left with speed 2420 m s –1. Oxygen molecule will move to the left with speed 136 m s –1. 5(a)(i) Both trolleys could not be stationary at the same time as this would imply that the total momentum of the system is zero at that instant. Since the total momentum of the system before collision was (1.5)(0.90) + (1.2)(-2.2) = -1.29 kg m s −1, with no external resultant force acting on the system, the total momentum should remain constant at this value throughout, according to the Principle of Conservation of Momentum. 5(a)(ii) According to Newton’s third law, the forces on trolleys A and B form an action- reaction pair. Hence, the average force between the trolleys