Question

A particle of mass 2 kg and velocity 5 m/s collides head on with another particle of mass 4 kg originally at rest. After the collision, the particle 1 continues in the same direction with the speed of 2.5 m/s. a. The velocity of particle 2 after collision is m/s. b. The energy (absolute value) lost in this collision is

Answer 1

Let the particle of mass m1 = 2kg

Having velocity V1 = 5 m/s

Now this particle collide with another particle having mass m2 = 4kg which is originally at rest i.e V2 = 0 m/s

Now after collision particle 1 continues in the same direction having velocity V3 = 2.5 m/s and particle 2 having velocity V4

Hence by the conservation of momentum

m1 *V1 + m2* V2 = m1*V3 + m2*V4

2*5 + 4*0 = 2*2.5 + 4* V4

V4 = (10 - 5 )/4

V4= 5/4

V4 = 1.25 m/s

This is the particle 2 velocity after the collision

b) now the energy lost in collision is =

(System has energy before collision) - (system has energy after collision )

Lost energy E = 0.5 * 2*5^2 - 0.5 *( 2* 2.5^2 + 4*1.25^2)

Lost energy E = 15.625 J

This much of energy gets loss during collision