Question

1. Two masses m and 2m are dropped from certain height h. then on reaching the ground
(a) K.E of them will be equal
(b) K.E of the heavier is 4 times the K.E of the lighter
(c) K.E of lighter is 4 times the K.E of the heavier
(d) K.E of the heavier is more than that of the lighter.

Answer 1

Answer:

(d) K.E of the heavier is more than that of the lighter.

Explanation:

Since the free falling acceleration is not dependent on mass, the final velocities of the both masses are same.

The kinetic energy

KE=mv^2/2

Hence, the kinetic energy of the heavier mass is two times of kinetic energy of lighter. Answer (d).

Answer 2

The kinetic energy of the heavier ball is twice that of the lighter ball.

Explanation:

If there are two bodies m and 2m. The kinetic energy of an object is given by :

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

For object 1,

$K_1=\dfrac{1}{2}mv^2$

For object 2,

$K_2=\dfrac{1}{2}(2m)v^2$

$K_2=2\times \dfrac{1}{2}mv^2$

$K_2=2\times K_1$

So, the kinetic energy of the heavier ball is twice that of the lighter ball. Hence, this is the required solution.

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Kinetic energy

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