Question

Case 1: A ball is made to strike a block at a speed of v m/s.
Case 2: The same ball is again made to strike the same block with the speed halved.
What will be the relation between the kinetic energies in both the cases?

• A. $2K{E}_1$ = $K{E}_2$
• B. $2K{E}_1$ = $2K{E}_2$
• C. $K{E}_1$ = $4K{E}_2$
• D. $4K{E}_1$ = $K{E}_2$

Answer 1

The correct option is C $K{E}_1$ = $4K{E}_2$

Kinetic energy of a body is given as,

K.E = $\frac{1}{2}m{v}^2$

Let,
Mass of body = m kg
Velocity of body = v m/s

Given:
Velocity in case 2 =$\frac{1}{2}$×Velocity in case 1

$v_2=\frac{1}{2}\times v_1$

Now, $K{E}_1=\frac{1}{2}m{v}_1^2$

and $K{E}_2=\frac{1}{2}m{v}_2^2$

Substituting the values for $v_1$ and $v_2$ in the equation for KE,

We get, $K{E}_2=\frac{1}{4}K{E}_1$

⇒ $4K{E}_2=K{E}_1$