Question

A body of mass $2.0kg$ makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed $\upsilon$ . What is the mass of the other body and the speed of the two body center of mass?

• A. 1.0 kg and $\frac{2}{3}v$
• B. 1.2 kg and $\frac{5}{8}v$
• C. 1.4 and $\frac{10}{17}v$
• D. 1.5 kg and $\frac{4}{7}v$

Answer 1

Answer: (B)

Given that $m_1=2kg{m}_2=?$
$u_1=v{v}_2=?$
$v_1=v/4v_{cm}=?$
From conservation of momentum
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
$2v+0=2\times \frac{v}{4}+m_2{v}_2$
$\frac{3}{2}v=m_2{v}_2\dots$ (i)
Now, from Newtons law of restitution
$v_1-v_2=e\left(u_2-u_1\right)$
$(\because e=1$ for elestic collision)
$\frac{v}{4}-v_2=-v{v}_2=\frac{5}{4}v$
From Eq. (i), we get
$m_2=\frac{6}{5}=1.2kg$
$v_{cm}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$
$=\frac{2\times \frac{v}{4}+1.2+\frac{5}{4}v}{2+12}$
$=\frac{(0.5+15)v}{32}=\frac{5}{8}v$