A pendulum, consisting of a ball of mass m on a light string of length 1.00 m, i

Question

A pendulum, consisting of a ball of mass m on a light string of length 1.00 m, is swung back to a 45° angle and released from rest. The ball swings down and, at its lowest point, collides with a block of mass 2m that is on a frictionless horizontal surface. After the collision, the block slides 1.00 m across the frictionless surface and an additional 0.370 m, before coming to rest, across a horizontal surface where the coefficient of friction between the block and the surface is 0.100. Use g = 9.80 m/s2.

(a) What is the block's speed after the collision?

(b) What is the velocity of the ball immediately after the collision? Use a plus sign if the velocity is in the same direction as the velocity of the ball just before the collision, and a negative sign if the velocity is in the opposite direction as the velocity of the ball just before the collision.

Explanation / Answer

Work done by the friction on the block = F x S

= (µmg)(S)

= (0.1 x 2m x 9.8) x 0.370

= 0.7252m

According to the law of conservation of the energy

Kinetic energy of the ball on the frictionless surface = Work done by the friction on the block

= 0.7252m

(0.5)(2m)v2 = 0.7252m

v2 = 0.7252

v = 0.8516 m/s

Conversing the energy between the point of release and the point of collision of the bob, we get

mgh = (0.5)(m)v2

v = (2gh)0.5

v = (2 x 9.8 x1(1-cos(45))0.5

V = 2.4 m/s

Applying law of conservation of momentum, we get

m x 2.4 = m x v + 2m x (0.8516)

v = 2.4 – 2 x 0.8516

   = 0.6968 m/s

Hence , block speed after collision = 0.8516 m/s

Velocity of the ball immediately after the collision = 0.6968 m/s

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