Question

A simple pendulum 2 m long has a bob of mass 0.5 kg which is 1 m above the floor at its

lowest point. The bob is pulled aside until it is 1.1 m above the floor. Its change in

potential energy is​

Answer 1

Answer:

change in potential energy is .05

Answer 2

Answer:

Velocity at lowest point of pendulum before collision

mgh=

2

1

mv

2

9.8×0.1=

2

1

×v

2

v

2

=1.96

v=1.4 m/s. i.e.,

momentum conservation

5×1.4=2.5×v

⇒v=2.8 m/s.