An object of mass m travelling with speed v has a head-on collision with another object of mass m travelling with speed v in the opposite direction.

Question 25

An object of mass m travelling with speed v has a head-on collision with another object of mass m travelling with speed v in the opposite direction. The two objects stick together after the collision.

 

What is the total loss of kinetic energy in the collision?

A 0                   B ½ mv²                         C mv²                                                 D 2mv²

 

 

 

Reference: Past Exam Paper – November 2015 Paper 12 Q13

 

 

 

Solution:

Answer: C.

Momentum is a vector quantity and we need to consider the direction.

Since the objects are moving in opposite direction, one of them will have a negative value.

Sum of momentum before collision = mv – mv = 0

 

Kinetic energy is a scalar quantity and so, the direction of motion does not matter.

Sum of KE before collision = ½ mv² + ½ mv² = mv²

 

Momentum is always conserved.

Sum of momentum after collision = Sum of momentum before collision = 0

Sum of momentum after collision = 0

2mv_f = 0

Giving the speed after collision, v_f = 0

That is, the objects do not move after the collision.

KE after collision = 0

 

Loss in KE = mv² – 0 = mv²