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

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

Question

(A) Given: $m_{1} = m$,$m_{2} = 2m$,$u_{1} = 2 \ m/s$,$u_{2} = 0$,and coefficient of restitution $e = 0.5$.Let $v_{1}$ and $v_{2}$ be the velocities a

Vedclass · Accepted Answer

$0, 1$

Vedclass · Answer

(A) Given: $m_{1} = m$,$m_{2} = 2m$,$u_{1} = 2 \ m/s$,$u_{2} = 0$,and coefficient of restitution $e = 0.5$.Let $v_{1}$ and $v_{2}$ be the velocities after collision.Applying the law of conservation of linear momentum:$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$$m(2) + 2m(0) = mv_{1} + 2mv_{2}$$2 = v_{1} + 2v_{2} \quad ...(i)$By the definition of the coefficient of restitution:$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$$0.5 = \frac{v_{2} - v_{1}}{2 - 0}$$1 = v_{2} - v_{1} \quad ...(ii)$Adding equations $(i)$ and $(ii)$:$(v_{1} + 2v_{2}) + (v_{2} - v_{1}) = 2 + 1$$3v_{2} = 3 \implies v_{2} = 1 \ m/s$Substituting $v_{2} = 1$ in equation $(ii)$:$1 = 1 - v_{1} \implies v_{1} = 0 \ m/s$Thus,the velocities after collision are $0 \ m/s$ and $1 \ m/s$.

$1$. Burning candle	$i$. Line spectrum
$2$. Sodium vapour	$ii$. Continuous spectrum
$3$. Bunsen flame	$iii$. Band spectrum
$4$. Dark lines in solar spectrum	$iv$. Absorption spectrum

$A$ 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

If $y = \sin(m \sin^{-1} x)$,then $(1 - x^2) y_2 - x y_1$ is equal to (Here,$y_n$ denotes $\frac{d^n y}{dx^n}$)

Along the $X$-axis,three charges $\frac{q}{2}, -q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2a$ respectively. The resultant electric potential at $x=a+r$ (if $a \ll r$) is: ($\varepsilon_0$ is the permittivity of free space)

If the points $(1,2)$ and $(3,4)$ lie on the same side of the straight line $3x - 5y + a = 0$,then $a$ lies in the set

$A$ prism of refractive index $\mu$ and angle $A$ is placed in the minimum deviation position. If the angle of minimum deviation is $A$, then the value of $A$ in terms of $\mu$ is :

Match the following:
$1$. Burning candle $i$. Line spectrum
$2$. Sodium vapour $ii$. Continuous spectrum
$3$. Bunsen flame $iii$. Band spectrum
$4$. Dark lines in solar spectrum $iv$. Absorption spectrum

Vedclass Test Series

Exam Paper Generator

Online Exam Module