Question

41. The bob (mass m) of a simple pendulum of length
L is held horizontal and then released. It collides
elastically with a block of equal mass lying on a
frictionless table. The kinetic energy of the block
will be :-
(1) Zero
(2) mgL
(3) 2mgl
mgL
(4)​

Answer 1

Answer:

(1) Zero

Explanation:

Answer 2

The kinetic energy of the block is $\dfrac{1}{2}mgl$

(4) $\dfrac{1}{2}mgl$

Explanation:

Given that,

Mass of bob = m

Length of pendulum = L

Mass of block = m

We need to calculate the velocity of bob

Using equation of motion

$v^2=u^2+2as$

put the value in the equation

$v=\sqrt{2gl}$

We need to calculate the velocity of the block

Using conservation of kinetic energy

$\dfrac{1}{2}m_{1}u_{1}^2+\dfrac{1}{2}m_{2}u_{2}^2=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

$\dfrac{1}{2}m_{1}u_{1}^2=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

$\dfrac{1}{2}m_{1}u_{1}^2=\dfrac{1}{2}(m_{1}+m_{2})v^2$

Here, $m_{1}=m_{2}=m$

$u_{1}=\sqrt{2gl}$

Put the value into the formula

$\dfrac{1}{2}m(\sqrt{2gl})^2=\dfrac{1}{2}\times 2m\times v^2$

$v^2=gl$

$v=\sqrt{gl}$

We need to calculate the kinetic energy of the block

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times m\times(\sqrt{gl})^2$

$K.E=\dfrac{1}{2}mgl$

Hence, The kinetic energy of the block is $\dfrac{1}{2}mgl$

Learn more :

Topic : Kinetic energy

https://brainly.in/question/7763241