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Question

Physics Question on work, energy and power

A point mass MM moving with a certain velocity collides with a stationary point mass M/2M/2. The collision is elastic and in one-dimension. Let the ratio of the final velocities of MM and M/2M/2 be xx. The value of xx is

A

22

B

33

C

1/21/2

D

1/41/4

Answer

1/41/4

Explanation

Solution

As collision is elastic both linear momentum and kinetic energy are conserved.
We have, following given condition,

Momentum conservation gives,
Mu1=Mv1+M2v2M u_{1}=M v_{1}+\frac{M}{2} v_{2}
2u1=2v1+v2(i)\Rightarrow 2 u_{1}=2 v_{1}+v_{2} \ldots (i)
Energy conservation gives,
12Mu12=12Mv12+12M2v22\frac{1}{2}Mu_{1}^{2} =\frac{1}{2}Mv_{1}^{2} + \frac{1}{2}\frac{M}{2} v_{2}^{2}
2u12=2u12+u22(ii)\Rightarrow\,\,\, 2u_{1}^{2} =2u_{1}^{2}+u_{2}^{2} \dots(ii)
Now, substituting for u1u_{1} from E(i)(i) in E (ii),(ii), we get
2(2u1+u22)2=2u12+u222\left(\frac{2u_{1}+u_{2}}{2}\right)^{2} =2u_{1}^{2}+u_{2}^{2}
4v12+v22+4v1v2=4v12+2v22\Rightarrow 4v_{1}^{2}+v_{2}^{2}+4v_{1}v_{2}=4v_{1}^{2}+2v_{2}^{2}
v22=4v1v2\Rightarrow v_{2}^{2}=4v_{1}v_{2}
or v1v2=14\frac{v_{1}}{v_{2}}=\frac{1}{4}
\therefore Ratio of final velocities of MM andM2\frac{M}{2} is 14 \frac{1}{4} .