Question

a 3.0 kg mass moving to the right at 1.4 m/s collides in a perfectly inelastic collision with a 2.0 kg mass initially at rest. what will the velocity of the combined mass be after the collision? show your work

Answer 1

The velocity of the combined mass after the collision is 0.84 ms-1.

Explanation:

According to law of conservation of momentum, the change in momentum before collision will be equal to the change in momentum of the objects after collision in isolated system.

But as it is perfectly inelastic collision in the present case, the final momentum will be based on the product of total mass of both the object with the velocity with which the collision occurred. This form is attained from the law of conservation of momentum as shown below:

So as law of conservation of momentum,

                   $M_{1} U_{1}+M_{2} U_{2}=M_{1} V_{1}+M_{2} V_{2}$

Here $M_{1}$ = 3 kg  and $M_{2}$ = 2 kg are the masses of objects 1 and 2, $U_{1}$ = 1.4 m/s  and $U_{2}$ = 0 are the initial velocities of object 1 and object 2,  $V_{1}$ and  $V_{2}$ are the final velocities of the objects.

So after collision, object 1 get sticked to object 2 and move together with equal velocity $V_{1}$ =  $V_{2}$ = $V_{f}$. Thus the above equation will become,

            $M_{1} U_{1}+M_{2} U_{2}=\left(M_{1}+M_{2}\right) V_{f}$

So the final velocity is

              $V_{f}=\frac{M_{1} U_{1}+M_{2} U_{2}}{\left(M_{1}+M_{2}\right)}$

Thus,

       $V_{f}=\frac{(3 \times 1.4+2 \times 0)}{(3+2)}=\frac{4.2}{5}$ = 0.84 ms-1.