TJC Unit 6 Collisions

Unit 06: Collisions ASSESSMENT OBJECTIVES Students should be able to (a) recall that impulse is given by the area under the force-time graph for a body and use this to solve problems. (b) state the principle of conservation of momentum. (c) apply the principle of conservation of momentum to solve simple problems including inelastic and (perfectly) elastic interactions between two bodies in one dimension (knowledge of the concept of coefficient of restitution is not required). (d) show an understanding that, for a (perfectly) elastic collision between two bodies, the relative speed of approach is equal to the relative speed of separation. (e) show an understanding that, whilst the momentum of a closed system is always conserved in interactions between bodies, some change in kinetic energy usually takes place.

Collisions occur when one object strikes another. It is the event in which two or more bodies exert forces on each other in about a relatively short time. 1.1 Impulse LO(a) Impulse is the product of the force acting on a body and the time interval during which the force is exerted. Consider a constant force F acting on an object for a time interval t as shown by the force–time graph in (a). The impulse is given by product of force and time interval = F t. (a) constant force (b) varying force In general, the force need not be constant, as shown in (b). Then impulse = dt F  = <F>t, where <F> is the average force. The average force <F> is defined as that constant force, which when acting over the same time interval t as the actual time-varying force, produces the same impulse and change in momentum. Graphically, the area under the average force–time graph is equal to the area under actual force– time graph. By Newton’s 2nd Law, dt dpF  => F t = p = mv The impulse of the force acting on a body is equal to the change in momentum of the body. Impulse is a vector quantity. Its unit is the same as momentum : kg m s-1 or N s. Example 1: Use the concept of impulse to explain: (a) "Follow through" when striking a tennis ball (b) Use of a seat belt in a car (c) Crumple zones of a car (a) The "follow through" increases the impact time of the racket and the ball. The impulse and hence the change in momentum is increased. Thus the ball goes off with a higher velocity. F/N Fmax <F> t/s t/s F/N F t t

(b) The seat belt extends slightly before arresting the forward lunge of the passenger. This increases the time at which the momentum of the passenger is brought to zero. The average force on the passenger is thus reduced. (c) The crumple zones of the car helps to increase the impact time as the momentum of the crashing car is brought to zero, thus reducing the average force on the passengers in the passenger compartment. Read more on crumple zone of car : Example 2: An object moving with an initial velocity of 12.0 m s-1 is acted on by a force for 22.0 s as shown in the graph below. (a) Calculate the change in momentum of the object, and (b) the average force <F> acting on the object. Example 3: A tennis ball of mass 0.060 kg and an initial speed of 30 m s-1 hits a racket. The ball rebounds from the bat with a speed of 40 m s-1 in the opposite direction. Given that the ball was in contact with the racket for 0.0020 s, calculate the average force on the tennis ball. s N 170 12.0) 0(10.0)(22. ½ graph time-force the under area (a)    dt Fp (b) Let <F> be the average force. <F> t = p <F> (22.0) = 170 => <F> =7.73 N 0 10.0 22.0 t/s F/N 10.0 <F>      left the to N101.2 3040060.00020.0 ][ 3   > F < > F < uvmp = t > F < u v <F>

1.2 Average force in multiple collisions Consider a body being hit N times in time interval t as shown by the force–time graph below. Then Impulse =  NptF  where t = time for N collisions Examples of multiple collisions: - a machine gun shooting a string of bullets, - gas molecules bombarding walls of a vessel. Example 4: A machine gun fires 50.0 g bullets at a speed of 1.00 x 103 m s-1. The gunner, holding the machine gun in his hands, can exert an average force of 180 N against the gun. Determine the maximum number of bullets he can fire per minute. How air bags works : 1.3 Principle of conservation of momentum LO(b) The principle of conservation of momentum states that the momentum of a system of objects remains constant if no resultant external force acts on the system. When two bodies collide, the forces F that they act on each other constitute an action-reaction pair. Therefore, each body receives an equal and opposite impulse, Ft, where t is the time of collision. As a result, the change in momentum of one body is equal and opposite to the other body. The total momentum of the two colliding bodies remains constant, hence the change in momentum of the system is zero. F/N Fmax N collisions <F> t/s t       11 3 min2166.3 1000.10500.0180     st N Nt NvmNptF <F> v

Proof: Consider an isolated system (one which has no external force acting on it) of two objects m1 and m2 whereby m1 is colliding with m2. Mathematical approach: If the two bodies are in contact for a time of t, From Newton's third law, F21 = –F12 F21t = –F12t --(1) impulse on m1 F21t = m1v1 – m1u1 --(2) impulse on m2 F12t = m2v2 – m2u2 --(3) Sub (2),(3) into (1) m1v1 – m1u1 = –(m2v2 – m2u2) => m1u1 + m2u2 = m1v1 + m2v2 Hence, the total momentum of the two objects before collision equals the total momentum after collision, which proves the principle of conservation of momentum. Graphical approach: The solid curve in the graphs below shows the variation with time of the force exerted by m1 on m2 during impact while the dotted curve shows the force that m2 exerts on m1. From Newton's third law, the two forces are equal and opposite at all instants. So the two curves are reflections of each other about the x-axis. Thus, area under F12 = – area under F21 impulse of m2 = – impulse of m1 change in momentum of m2 = – change in momentum of m1 => change in momentum of m2 + change in momentum of m1 = 0 F12 F21 area = impulse of m2 area = impulse of m1 F/N t/s m1 u1 before collision u2 m2 m1 v1 after collision v2 m2 during impact m1 m2 F12 F21

The result shows that the change in momentum of the system is zero, which again proves the principle of conservation of momentum. Note:  F12 and F21 are internal forces of the system. i.e. forces which arise between the interacting objects of the system.  If there is an external force, for example, friction, then there will be an overall change in the momentum of the system caused by the external force. 1.4 Types of collisions LO(d)(e) Generally, there are 3 types of collisions:  Elastic collision in which both momentum and kinetic energy are conserved.  Inelastic collision in which momentum is conserved but kinetic energy is not.  Completely inelastic collision in which momentum is conserved and the particles stick together after collision so that their f