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Question

Physics Question on collision theory

A body of mass 5kg5 \,kg makes an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal to 110\frac{1}{10} th of its original velocity. Then the mass of the second body is

A

4.09 kg

B

0.5 kg

C

5 kg

D

5.09 kg

Answer

4.09 kg

Explanation

Solution

Mass of the first body m1=5m_1 = 5 kg and for elastic collision coefficient of restitution, e=1.e = 1.

Let initially body m1m_1 moves with velocity v after collision velocity becomes (u10)\left( \frac{u}{10}\right) .
Let after collision velocity of M block becomes (v2)(v_2) .
By conservation of momentum
m1u1+m2u2=m1v1+m2v2m_1u_1 + m_2 u_2 = m_1v_1 + m_2 v_2
or 5u+M×0=5×u10+Mv25u + M\times 0 = 5 \times \frac{u}{10} + Mv_2
or 5u=u3+Mv25u = \frac{u}{3} + Mv_2 ....(i)
Since , v1v2=e(u1u2)v_1 - v_2 = - e(u_1 - u_2)
or u10v2=1(u)\frac{u}{10} - v_2 = - 1(u) or u10+u=v2\frac{u}{10} + u = v_2
or 11u10=v2\frac{11 u}{10} =v_2 ......(ii)
Substituting value of v2v_2 in E (i) from E (ii)
5u=u2+M(11u10)5u = \frac{u}{2} +M \left(\frac{11u}{10}\right)
or 512=M(1110) 5 - \frac{1}{2} = M \left(\frac{11}{10}\right)
M=9×102×11=4511\Rightarrow M = \frac{9 \times10}{2 \times11} = \frac{45}{11}
=4.09kg= 4.09 \,kg