Q#772: [Dynamics > Collisions](Past Exam Paper – June 2011 Paper 12 Q13)
Diagram shows a particle P, travelling at speed v, about to collide with a stationary particle Q of the same mass. The collision is perfectly elastic.
Which statement describes motion of P and of Q immediately after the collision?
A P rebounds with speed ½ v and Q acquires speed ½ v.
B P rebounds with speed v and Q remains stationary.
C P and Q both travel in the same direction with speed ½ v.
D P comes to a standstill and Q acquires speed v.
Solution 772:
Answer: D.
For a perfectly elastic collision, both momentum and energy are conserved.
From the law of conservation of momentum,
Sum of momentum before collision = Sum of collision after collision
Before collision,
Sum of momentum, p = mv + 0 = mv
Total kinetic energy = ½ mv2
After collision,
Consider choice A,
Sum of momentum, p = m(0.5v) + m(-0.5v) = 0. [A is incorrect]
Consider choice B,
Sum of momentum, p = - mv [B is incorrect]
Consider choice C,
Sum of momentum, p = m(0.5v) + m(0.5v) = mv
Total kinetic energy = ½ m (0.5v)2 + ½ m (0.5v)2 = 0.25mv2 [C is incorrect]
Consider choice D,
Sum of momentum, p = m(0) + mv = mv
Total kinetic energy = ½ mv2
Both momentum and energy are conserved.
Post a Comment for "Diagram shows a particle P, travelling at speed v, about to collide with a stationary"