Question

A $50$ kg mass is travelling at a speed of $2$ m/s. Another $60$kg mass is travelling at a speed of $12$ m/s in the same direction, strikes the first mass. After the collision the $50$kg mass is travelling with a speed of $4$ m/s. The coefficient of restitution of the collision is :

• A. $\frac{19}{30}$
• B. $\frac{30}{19}$
• C. $\frac{20}{11}$
• D. $\frac{11}{20}$

Answer 1

The correct option is A $\frac{19}{30}$

Given,

$m_1=50kg$

$m_2=60kg$

$v_{1i}=2m/s$

$v_{2i}=12m/s$

$v_{1f}=4m/s$

We know that there is no external force on the system of two mass.

So here we can use momentum conservation for the collision.

$m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f}$

Now we have,

$50\left(2\right)+60\left(2\right)=50\left(4\right)+60v_{2f}$

$100+720=200+60v$

$620=60v$

$v_{2f}=\frac{31}{3}m/s$

Now by formula of coefficient of restitution, we will have,

$e=\frac{v_{2f}-v_{1f}}{v_{1i}-v_{2i}}$

Substituting the values, we get,

$e=\frac{\frac{31}{3}-4}{12-2}=\frac{19}{30}$