A pendulum, consisting of a ball of mass m on a light string of length 1.00 m, i
ID: 1366448 • Letter: A
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 sided 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 0.130. Use g = 9.80 m/s^2.Explanation / Answer
distance travelled on frictional surface = x = 0.4 m
fricitional force fk = uk*m2*g
work done = fk*x
inital KE = KE1 = 0.5*m2*v2^2
final KE = 0
from work energy theorem
work done = change in KE
uk*m2*g*x = 0.5m2v^2
0.13*9.8*0.4 = 0.5*v2^2
v2 = 1.009 m/s <<--------answer
(b)
initial PE of the ball = PE1 = m1*g*L*(1-cos45)
at the lowest point KE = 0.5*m1*u^2
from energy conservation
KE = PE
u = sqrt(2*g*L*(1-cos345))
u = sqrt(2*9.8*1*(1-cos45)) = 2.4 m/s
from momentum conservation
momentum before colision = momentum after collsion
m1*u = m1*v1 + m2*v2
m*2.4 = m*1.009 + 2m*v2
v2 = 1.391 m/s <,----answer
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