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 degree 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 0.350 m, before coming to rest, across a horizontal surface where the coefficient of friction between the block and the surface is 0.110. Use g = 9.80 m/s^2. What is the block's speed after the collision? 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

a) length of string= 1m ,

mass of pendulum = m

the is raised to 45 degrees , so , perpendicular height raised = (1 - 1* cos(45) = 0.293 m

now, the potential energy is converted to kinetic enrgy at the lowest point,

so , mgh = 0.5* m* v2

=> v = sqrt ( 2gh)

=> V= 2.396m/s

a) let the velocity of the block after hitting on the frictionless surface be Vf

and final velocity be zero

now , it stops over a distance of 0.35 m

where frictional deacceleration = -2.156 m/s

so, we have ,

02-Vf2 = 2* (-2.156 )* ( 0.35 m)

=>vf= 1.228 m/s

b) now, use conservation of momentum

inital momentum = final momentum

momentum of ball = momentum of block + momentum of ball

m* V= 2m* Vf + m* Vball

=> 2.396 = 2* 1.228 + Vball

=>Vball = -0.06 m/s ( negative sign indicates that the ball moves opposite to the direction of motion of block )

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