A ball moving with velocity 2, m/s collides h | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Here, $m_{1}=m, m_{2}=2 m$

$\mathrm{u}_{1}=2 \mathrm{m} / \mathrm{s}, \mathrm{u}_{2}=0$

Coefficient of restitution, $\mathrm{e}=0.5$

Let $v_{

Vedclass · Accepted Answer

$0, 1$

Vedclass · Answer

Here, $m_{1}=m, m_{2}=2 m$

$\mathrm{u}_{1}=2 \mathrm{m} / \mathrm{s}, \mathrm{u}_{2}=0$

Coefficient of restitution, $\mathrm{e}=0.5$

Let $v_{1}$ and $v_{2}$ be their respective velocities afte collision.

Applying the law of conservation of linea momentum, we get,

${\mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}=\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}}$

${\mathrm{m} 	imes 2+2 \mathrm{m} 	imes 0=\mathrm{m} 	imes \mathrm{v}_{1}+2 \mathrm{m} 	imes \mathrm{v}_{2}}$

or $2 \mathrm{m}=\mathrm{mv}_{1}+2 \mathrm{mv}_{2}$

or $2=\left(v_{1}+2 v_{2}ight)$        $...(i)$

By definition of coefficient of restitution,

${\mathrm{e}=\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{u}_{1}-\mathrm{u}_{2}}}$

Or     ${\mathrm{e}\left(\mathrm{u}_{1}-\mathrm{u}_{2}ight)=\left(\mathrm{v}_{2}-\mathrm{v}_{1}ight)}$

$0.5(2-0)=\left(v_{2}-v_{1}ight)$         $...(ii)$

$1=v_{2}-v_{1}$

Solving equations $(i)$ and $(ii),$ we get,

$\mathrm{v}_{1}=0 \mathrm{m} / \mathrm{s}, \mathrm{v}_{2}=1 \mathrm{m} / \mathrm{s}$

A ball moving with velocity $2\, m/s$ collides head-on with another stationary ball of double the mass. If the coefficient of restitution is $0.5$, then their velocities (in $m/s$) after collision will be

Similar Questions

A ball of mass $m$ is dropped from a heigh $h$ on a platform fixed at the top of a vertical spring, as shown in figure. The platform is depressed by a distance $x.$ Then the spring constant is

Two blocks $A$ and $B$ of masses $1\,\,kg$ and $2\,\,kg$ are connected together by a spring and are resting on a horizontal surface. The blocks are pulled apart so as to stretch the spring and then released. The ratio of $K.E.s$ of both the blocks is

An object has momentum $p$ & kinetic energy $E$. If its momentum becomes $2\,p$ then its kinetic energy will be :-

The potential energy of a diatomic molecule is given by $U = \frac{A}{{{r^{12}}}} - \frac{B}{{{r^6}}}$ . $A$ and $B$ are positive constants. The distance $r$ between them at equilibrium is