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 0.410 m, before coming to rest, across a horizontal surface where the coefficient of friction between the block and the surface is 0.140. Use g = 9.80 m/s2. 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

acceleration of block on rough surface = - uk.g = - (0.140 x 9.81) = - 1.37 m/s^2

and block comes to rest in 0.410 m.

using v^2 - u^2 = 2 a d

0 - u^2 =2 x -1.37 x 0.410

u = 1.06 m/s ......speed of 2m block just after collision,

for ball, mgh = mv^2 /2

v = sqrt(2 x 9.80 x (1 - cos45)) = 2.40 m/s speed of ball just before collision towwards 2m nlock .

using momentum conservation,

m ( 2.40) - ( 2m x 0) = (2m x 1.06) + mvf

vf = 0.28 m/s .......Ans

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