Question

A rubber ball of mass 1 kg hits the ground at an angle of 30º with ground and rebounds at angle of 60º with ground without change in speed. If the speed of ball is 1 m/s, then impulse imparted by ground to ball, is

Answer 1

The impulse imparted by ground to ball is (√3 + 1)/2 kg-m/s

Explanation:

Given:

Mass of the rubber ball m = 1 kg

Angle of impact $\theta_1$ = 30º

Angle of rebound $\theta_2$ = 60º

Initial speed u = 1 m/s

To find out:

Impulse imparted by the ground to the ball

Solution:

Initial vertical component of the velocity

$u_y=u\sin\theta_1$

$\implies u_y=1\times\sin 30^\circ$

$\implies u_y=\frac{1}{2}$ m/s

Since the speed does not change after impact

Therefore, final rebound speed v = 1 m/s

Final vertical component of the velocity

$v_y=-v\sin\theta_2$

$\implies v_y=-1\times\sin 60^\circ$

$\implies v_y=\frac{\sqrt{3}}{2}$ m/s

Impulse imparted by the ground = Change in momentum

$\implies F\Delta t=m(v_y-u_y)$

$\implies F\Delta t=1(-\frac{\sqrt{3}}{2}-\frac{1}{2})$

$\implies F\Delta t=-\frac{(\sqrt{3}+1)}{2}$ kg-m/s

Since the impulse is imparted by the ground in the upward direction

Therefore, it is of negative sign

Hope this answer is helpful.

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