Question

A body of mass 8 kg moving with velocity 14 m/s collides
with another body of mass 7 kg at rest. If two bodies
stick together after collision, then the loss of kinetic
energy of system is​

Answer 1

Answer:

hui bye

Explanation:

Answer 2

Answer:

A body of mass 8 kg is moving with velocity 14 m/s collides  with another body of mass 7 kg which is at rest, after collision two bodies  stick together and moves, then the loss in kinetic  energy of the system is $373.3J$

Explanation:

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

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

it can be expressed as

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

mass of the first body $m_{1} =8kg$

the velocity of the first body $u_{1} =14m/s$

mass of the second body $m_{2} =7kg$

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

substitute these values to get the velocity of the combined body

$8*14+0=(8+7)V\\V=7.4m/s$

The kinetic energy before the collision

 $KE_{1}=\frac{1}{2} m_{1} u_{1} ^{2}\\ =\frac{8*169}{2}=784J$

the kinetic energy after the collision $KE_{2}=\frac{1}{2}(m_{1} +m_{2} )V^{2}$

$=\frac{15*7.4^{2} }{2}=410.7J$

therefore the loss in kinetic energy $784-410.7=373.3J$