Question

A body of mass $3kg$ collides elastically with another body at rest and then continues to move in the original direction with one half of its original speed. What is the mass of the target body?
$A)\text{ }1kg$
$B)\text{ 2}\text{.5}kg$
$C)\text{ 2}kg$
$D)\text{ 5}kg$

Answer 1

Hint: This problem can be solved by using the direct formula for the final velocity of a body participating in an elastic collision with another body in terms of the masses of the body and their initial velocities. Then by plugging in the final velocity of the body as given in the question, we can get the required answer.

Formula used:
$v=\dfrac{m-M}{m+M}u+\dfrac{2M}{m+M}u'$

Complete step by step answer:
We shall use the direct formula for the final velocity of a body after an elastic collision with another body.
The final velocity $v$ of a body of mass $m$ after an elastic collision with another body of mass $M$ is given by
$v=\dfrac{m-M}{m+M}u+\dfrac{2M}{m+M}u'$ --(1)
Where $u,u'$ are the initial velocities of the bodies of mass $m$ and $M$ respectively before the collision.
Now, let us analyze the question.
The mass of one of the bodies participating in the elastic collision is $m=3kg$.
The initial velocity of the body is $u$.
Let the final velocity of the body be after the elastic collision be $v$.
Let the mass of the other body participating in the elastic collision be $M$.
It is given that the other body was at rest before the collision. Therefore, the initial velocity of this body is $u'=0$.
Now, using (1), we get
$v=\dfrac{m-M}{m+M}u+\dfrac{2M}{m+M}u'$
Putting the values in the above equation, we get
$v=\dfrac{3-M}{3+M}u+\dfrac{2M}{3+M}\left( 0 \right)=\dfrac{3-M}{3+M}u+0=\dfrac{3-M}{3+M}u$ --(2)
Now, according to the question, the body of mass $m$ moves in the same direction with half of its original speed after the collision. Therefore,
$v=\dfrac{1}{2}u$ --(3)
Putting (3) in (2), we get
$\dfrac{1}{2}u=\dfrac{3-M}{3+M}u$
$\Rightarrow \dfrac{3-M}{3+M}=\dfrac{1}{2}$
$\Rightarrow 2\left( 3-M \right)=3+M$
$\Rightarrow 6-2M=3+M$
$\Rightarrow M+2M=6-3=3$
$\Rightarrow 3M=3$
$\Rightarrow M=\dfrac{3}{3}=1kg$
Hence, the mass of the second body is $1kg$.
Therefore, the correct option is $A)\text{ }1kg$.

Note: Students must remember that formula (1) is valid only for perfectly elastic collisions (coefficient of restitution $e=1$). This is because in elastic collisions, both the total linear momentum as well as the total kinetic energy of the system is conserved. This gives rise to two equations which can be solved simultaneously to get the direct formulae for the final velocities of the bodies after the elastic collision. Students must not make the mistake of using these formulas for inelastic collisions.