First slide

Kinetic energy

Question

A running man has half the kinetic energy than a boy of half his mass has. The man speeds up by I m/sec and then has the same kinetic energy as the boy. What were the original speeds of man and boy respectively?

Moderate

A

2.4, 4.8 ${ms}^{- 1}$
B

$2.4, 3.4 {ms}^{- 1}$
C

$3.4, 4.8 {ms}^{- 1}$
D

$3.4, 6.8 {ms}^{- 1}$

Solution

Let, M = mass of man, m = mass of boy

V = speed of man, v = speed of boy

Given: $\frac{1}{2} {MV}^{2} = \frac{1}{2} (\frac{1}{2} {mv}^{2}), As m = \frac{M}{2}$

So $\frac{1}{2} {MV}^{2} = \frac{1}{2} (\frac{1}{2} \times \frac{M}{2} v^{2}),$
Hence, $v^{2} = 4 V^{2} or v = 2 V$
When the man speeds up by I m./s, then we get

$\frac{1}{2} M {(V + 1)}^{2} = \frac{1}{2} {mv}^{2} = \frac{1}{2} \frac{M}{2} (4 V^{2})$
or ${(V + 1)}^{2} = 2 V^{2} or V^{2} - 2 V - 1 = 0$
Solving we get: V:2.4 m/s and v : 4.8 m/s

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