Question

A pendulum with a bob of mass 30g is lifted to height of 5.00cm above its lowest point and allowed to fall so that collides elastically with a block of mass 20g that rests on a smooth horizontal floor. The block stops 5 seconds after the collision. Assuming no sound is generated during the collision, determine the:
a)speed of the bob just before it collides with the block
b)distance travelled by the block before it stops
c)deceleration of the block

Answer 1

PE=KE
m₁gh =m₁v₁²/2
v₁ =sqrt(2gh) = sqrt(2•0.03•0.05) =
=0.055 m/s.

m₁v₁ = m₁u₁ + m₂u₂
u₂=2m₁v₁/(m₁+m₂)=2•0.03•0.055/(0.03+0.02) = 0.066 m/s.

0= u₂ -at
a= u₂/t=0.066/5= 0.0132 m/s²,

s=at²/2=0.0132•5²/2=0.165 m