Question 317: Centre of Mass and Rotational Motion

Question

A moving body with mass ' $\mathrm{m}_1$ ' strikes a stationary mass ' $\mathrm{m}_2$ '. What should be the ratio $\frac{m_1}{m_2}$ so as to decrease the velocity of first by (1.5) times the velocity after the collision?

Accepted Answer

Let initial velocity of mass $m_1$ be $v_1$ and final velocity of mass $m_2$ be $v_2$
According to the given condition,
Final velocity of mass $m_1$ is $\frac{v_1}{1.5}=\frac{2}{3} v_1$
Coefficient of restitution,
$$\begin{aligned}
& e=\frac{	ext { Velocity after collision }}{	ext { Velocity beforecollision }} \
1 & =\frac{\left(\mathrm{v}_2-\frac{2 \mathrm{v}_1}{3}ight)}{\mathrm{v}_1} \quad \ldots(\mathrm{e}=1, 	ext { for elastic collision }) \
	herefore \quad \mathrm{v}_2 & =\frac{5 \mathrm{v}_1}{3}\quad	ext{.... (i)}
\end{aligned}$$
By following conservation of momentum
$$\begin{aligned}
& \mathrm{m}_1 \mathrm{v}_1=\frac{\mathrm{m}_1 \mathrm{v}_1}{1.5}+\mathrm{m}_2 \mathrm{v}_2 \
& \mathrm{~m}_1 \mathrm{v}_1=\mathrm{m}_1 \frac{2}{3} \mathrm{v}_1+\mathrm{m}_2 \frac{5}{3} \mathrm{v}_1 \
& \frac{1}{3} \mathrm{~m}_1 \mathrm{v}_1=\mathrm{m}_2 \frac{5}{3} \mathrm{v}_1 \Rightarrow \frac{\mathrm{~m}_1}{\mathrm{~m}_2}=\frac{5}{1}
\end{aligned}$$

Answer

$1: 25$

Answer

$1: 5$

Answer

$5: 1$

Answer

$25: 1$

Centre of Mass and Rotational Motion

A moving body with mass ' \mathrm{m}_1 ' strikes a stationary mass ' \mathrm{m}_2 '. What should be the ratio \frac{m_1}{m_2} so as to decrease the velocity of first by (1.5) times the velocity after the collision?

Choose the correct answer: