A ball moving with velocity 2 , m/s collides | 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 $e = 0.5$.Applying the law of conservation of linear momentum:$m_1 u_1 + m_2 u_2 = m_1

Vedclass · Accepted Answer

$0, 1$

Vedclass · Answer

(A) Given: $m_1 = m$,$m_2 = 2m$,$u_1 = 2 \, m/s$,$u_2 = 0$,and $e = 0.5$.Applying the law of conservation of linear momentum:$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$m(2) + 2m(0) = m v_1 + 2m v_2$$2m = m v_1 + 2m v_2 \implies v_1 + 2v_2 = 2$ $...(i)$Using the definition of the coefficient of restitution:$e = \frac{v_2 - v_1}{u_1 - u_2}$$0.5 = \frac{v_2 - v_1}{2 - 0}$$v_2 - v_1 = 1$ $...(ii)$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 $v_2 = 1$ into equation $(ii)$:$1 - v_1 = 1 \implies v_1 = 0 \, m/s$Thus,the velocities after collision are $0 \, m/s$ and $1 \, m/s$.

$A$ 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

We have two spheres,one of which is hollow and the other solid. They have identical masses and moments of inertia about their respective diameters. The ratio of their radii is given by

Magnetic lines of force due to a bar magnet do not intersect because

If $ax^2+2hxy-2ay^2+3x+15y-9=0$ represents a pair of lines intersecting at $(1,1)$,then $ah=$

The magnetic field at the centre of a circular current-carrying conductor of radius $r$ is $B_c$. The magnetic field on its axis at a distance $r$ from the centre is $B_a$. The value of $B_c : B_a$ will be:

The temperature in $K$ at which $\Delta G = 0$ for a given reaction with $\Delta H = -20.5 \ kJ \ mol^{-1}$ and $\Delta S = -50.0 \ J \ K^{-1} \ mol^{-1}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module