A body of mass 2 kg moving with a speed of 4 m/s. makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial speed. The speed of the two body centre of mass is $\frac{x}{10} m / s$ . Then the value of $x$ is_________.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 25.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

From law of conservation of linear momentum
$P_{i} = P_{f} \Rightarrow m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}$ …………….. (i)
$2 \times 4 + 0 = 2 \times 1 + m_{2} v_{2}$
In elastic collision coefficient of restitution, $e = 1$ and $e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$
$\Rightarrow v_{2} - v_{1} = e (u_{1} - u_{2}) \Rightarrow v_{2} - 1 = 1 (4 - 0) o r, v_{2} = 5$
Now from eq. (i)
$8 = 2 + m_{2} \times 5 \Rightarrow m_{2} = \frac{6}{5}$
Velocity of centre of mass
$V_{c m} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}} = \frac{2 \times 4 + 0}{2 + \frac{6}{5}} = \frac{25}{10} m / s$