A body mass of 1 kg, initially at rest explodes and breaks into three parts. The masses of the parts are in the ratio 1: 1: 3 . The two pieces of equal mass fly off perpendicular to each other with a speed of 30 m / s each. The velocity of the heavier part in m / s is

Question

A body mass of 1 kg, initially at rest explodes and breaks into three parts. The masses of the parts are in the ratio 1: 1: 3 . The two pieces of equal mass fly off perpendicular to each other with a speed of 30 m / s each. The velocity of the heavier part in m / s is (A) 10 √2 (B) 6 (C) 3 (D) 6 √2

Tardigrade Admin · Accepted Answer

Initial momentum before explosion is zero. Final momentum after explosion.

2 \frac{m}{4} \times 30 - \frac{3 m}{4} v^{'}

From conservation of momentum
Initial momentum = Final momentum

\Rightarrow 2 \frac{m}{4} \times 30 - \frac{3 m}{4} v^{'} = 0

\Rightarrow v^{'} = 102 m / s

Q. A body mass of $1 k g$ , initially at rest explodes and breaks into three parts. The masses of the parts are in the ratio $1 : 1 : 3.$ The two pieces of equal mass fly off perpendicular to each other with a speed of $30 m / s$ each. The velocity of the heavier part in $m / s$ is

$102$

$6$

$3$

$62$

Initial momentum before explosion is zero. Final momentum after explosion.

$2 \frac{m}{4} \times 30 - \frac{3 m}{4} v^{'}$
From conservation of momentum
Initial momentum = Final momentum
$\Rightarrow 2 \frac{m}{4} \times 30 - \frac{3 m}{4} v^{'} = 0$
$\Rightarrow v^{'} = 102 m / s$

Q. A body mass of 1kg, initially at rest explodes and breaks into three parts. The masses of the parts are in the ratio 1:1:3. The two pieces of equal mass fly off perpendicular to each other with a speed of 30m/s each. The velocity of the heavier part in m/s is

102​

6

3

62​

Initial momentum before explosion is zero. Final momentum after explosion. 2​4m​×30−43m​v′ From conservation of momentum Initial momentum = Final momentum ⇒2​4m​×30−43m​v′=0 ⇒v′=102​m/s

Q. A body mass of $1 k g$ , initially at rest explodes and breaks into three parts. The masses of the parts are in the ratio $1 : 1 : 3.$ The two pieces of equal mass fly off perpendicular to each other with a speed of $30 m / s$ each. The velocity of the heavier part in $m / s$ is

$102$

$6$

$3$

$62$

Initial momentum before explosion is zero. Final momentum after explosion.

$2 \frac{m}{4} \times 30 - \frac{3 m}{4} v^{'}$
From conservation of momentum
Initial momentum = Final momentum
$\Rightarrow 2 \frac{m}{4} \times 30 - \frac{3 m}{4} v^{'} = 0$
$\Rightarrow v^{'} = 102 m / s$