Question

A 2.5 kg body moving with velocity 6 m/s in a straight line collides a 1.5 kg body initially at rest. After collision two bodies stick together. Find the speed of the combined body and loss in kinetic energy​

Answer 1

Answer:

A 2.5 kg body moving with a velocity of 6 m/s in a straight line collides a 1.5 kg body initially at rest. After the collision, two bodies stick together, the speed of the combined body is $1.5m/s$ and loss in kinetic energy​ is $40.5J$

Explanation:

In a collision, momentum should be conserved which means total momentum after the collision is equal to the total momentum before the collision

after the collision, two bodies stick together and move with velocity $V$

it can be expressed as

$m_{1} u_{1} +m_{2} u_{2} =(m_{1} +m_{2} )V$

from the question, it is given that

mass of the first body $m_{1} =2.5kg$

the velocity of the first body $u_{1} =6m/s$

mass of the second body $m_{2} =1.5kg$

the second body is initially at rest so the velocity $u_{2} =0$

substitute these values to get the velocity of the combined body

$2.5*6+0=(2.5+1.5)V\\V=6/4=1.5m/s$

The kinetic energy before the collision is

$KE=\frac{1}{2}m_{1}u_{1} ^{2} =\frac{2.5*36}{2}=45J$

the kinetic energy after the collision

$KE=\frac{1}{2}(m_{1} +m_{2} )V^{2}$

$=\frac{4*2.25}{2}=4.5J$

the loss in kinetic energy is $45-4.5=40.5J$