A physics student is playing a game in which throws a ball (mass 0.43 kg ) at a

Question

A physics student is playing a game in which throws a ball (mass   0.43 kg   ) at a target (mass   1.83 kg   ), which is hanging from a massless wire of length   81 cm   . The ball collides elastically with the target, which then swings backwards. We consider a case in which the ball was moving horizontally with a velocity   2.19 m/s    right before the collision.

Part A-What is the velocity of the ball right after the collision?

Part B-What is the velocity of the target right after the collision?

Part C-What is the maximum angle (with vertical) the string makes as the target rises?

Part D-If the collision occured over a time interval of   0.45

s   , what is the average force exerted on the target by the ball?

Explanation / Answer

A) velocity of the ball right after the collison

v1 = (m1-m2)*u1/(m1+m2) = (0.43-1.83)*2.19/(0.43+1.83) = -1.35 m/s

the ball moves with 1.35 m/s but opposite to the original direction of the ball

B) v2 = 2*m1*u1/(m1+m2) = (2*0.43*2.19)/(0.43+1.83) = 0.83 m/s

C) using conservation of energy

energy of the target right after collision = energy at the extreme position of the retrn point

0.5*m2*v2^2 = m2*g*h

h = (0.5*0.83*0.83)/(9.81) = 0.035 m

but height from the ground is h = l*(1-cos(theta))

1-cos(theta) = h/l = 0.035/0.81 = 0.0432

cos(theta) = 1-0.0432 = 0.9568

theta = acos(0.9568) = 16.9 degrees

D) by impulse momentumtheorem

Favg*dt = m2*(v2-u2) = 1.83*(0.83)

Favg = 1.5189/0.45 = 3.37 N

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