A ball of mass m moving with speed u collides | vedclass

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

Question

(D) The velocity of the ball before collision can be resolved into two components: horizontal component $u_x = u \cos 	heta$ and vertical component $

Vedclass · Accepted Answer

$m u(1+e) \sin 	heta$

Vedclass · Answer

(D) The velocity of the ball before collision can be resolved into two components: horizontal component $u_x = u \cos 	heta$ and vertical component $u_y = -u \sin 	heta$ (taking downward as negative).Since the surface is smooth,there is no impulse in the horizontal direction,so the horizontal velocity remains unchanged: $v_x = u \cos 	heta$.In the vertical direction,the coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach along the normal: $e = \frac{v_y}{u_y}$.Thus,the vertical velocity after collision is $v_y = e u \sin 	heta$ (upward).The impulse $I$ imparted to the ball by the surface is equal to the change in momentum of the ball: $\vec{I} = m(\vec{v} - \vec{u})$.Since there is no change in horizontal momentum,the impulse is only in the vertical direction: $I = m(v_y - u_y) = m(e u \sin 	heta - (-u \sin 	heta)) = m u(1+e) \sin 	heta$.The magnitude of the impulse imparted to the surface by the ball is equal to the magnitude of the impulse imparted to the ball by the surface,which is $m u(1+e) \sin 	heta$.

$A$ ball of mass $m$ moving with speed $u$ collides with a smooth horizontal surface at an angle $\theta$ with it as shown in the figure. The magnitude of the impulse imparted to the surface by the ball is [Coefficient of restitution of collision is $e$].

Similar Questions

$A$ body of mass $1\,kg$ collides head-on elastically with a stationary body of mass $3\,kg$. After the collision,the smaller body reverses its direction of motion and moves with a speed of $2\,m/s$. The initial speed of the smaller body before the collision is $..........\,m/s$.

$A$ light particle moving horizontally with a speed $v_1$ strikes a very heavy block moving in the same direction with a speed $v_2$. The collision is elastic. After the collision,the velocity of the particle is:

$A$ sphere of mass $0.1 \ kg$ is attached to a string of length $1 \ m$. It is released from a horizontal position and collides with another sphere of equal mass $m$ at rest. Find the kinetic energy gained by the second sphere. The collision is perfectly elastic. (in $J$)

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest,the fractional loss of its energy is $P_d$; while for its similar collision with a carbon nucleus at rest,the fractional loss of energy is $P_c$. The values of $P_d$ and $P_c$ are respectively:

$A$ body of mass $2\,kg$ makes an elastic collision with a second body of mass $m$ at rest and continues to move in the original direction but with one-fourth of its original speed. What is the mass of the second body? ................ $kg$

Vedclass Test Series

Exam Paper Generator

Online Exam Module