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 2mthat is on a frictionless horizontal surface.

Note that you will most likely get different numbers for the different parts below. As usual, a good way to handle that is to solve the problem using variables, and only plug in numbers at the end.

a. After the collision, the block slides 1.00 m across the frictionless surface and an additional 0.400 m, before coming to rest, across a horizontal surface where the coefficient of friction between the block and the surface is 0.150. Use .

Calculate the speed the block has immediately after the collision .

__1.08_____ m/s

b. After the collision, the block slides 1.00 m across the frictionless surface and an additional 0.400 m, before coming to rest, across a horizontal surface where the coefficient of friction between the block and the surface is 0.150. Use .

Calculate 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.
_______ m/s

HELP WITH B

Answer 1

(al Using v²_u = zad o-u2=2(-ugla tu²=42+0.1579.87 0.4 u= 1.084mla

(B) First finding velocity of ball before collision using conservation of energy I na U2 = mylli-cose) tu? = 9.8(1)(i-Cos450) = U,= 2.396 mb Now using conservation of momentum my = myut mir w(2.396) = (2 24) (1084) + alv.) v=[0.228 mb

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