Question

A 1kg stationary bomb is exploded in three parts having masses in the ratio 1:1:3 respectively. Parts having same mass move in perpendicular direction with velocity 30m/s, then the velocity of bigger part will be?
Ans : 14.14 m/s

Answer 1

Answer on Question #51607 - Physics - Mechanics | Kinematics | Dynamics

A 1 kg stationary bomb is exploded in three parts having masses in the ratio 1:1:3 respectively. Parts having same mass move in perpendicular direction with velocity $30\mathrm{m/s}$, then the velocity of bigger part will be?

Solution

$$m_s = 1\ \mathrm{kg};\ \mathrm{m}_1:\mathrm{m}_2:\mathrm{m}_3 = 1:1:3;\ \mathrm{v}_1 = \mathrm{v}_2 = 30\ \mathrm{m/s}.$$ $$\mathrm{m}_1 = \mathrm{m}_2 = \mathrm{m} = 1/5\ \mathrm{kg} = 0.2\ \mathrm{kg}.$$ $$\mathrm{m}_3 = 3\mathrm{m} = 0.6\ \mathrm{kg}.$$

The Law of conservation of momentum:

$$\mathrm{OX:}\ 0 = \mathrm{m}_1\mathrm{v}_1 - \mathrm{m}_3\mathrm{v}_{\mathrm{x}3};\ \mathrm{mv}_1 = 3\mathrm{mv}_{\mathrm{x}3};\ \mathrm{v}_{\mathrm{x}3} = \mathrm{v}_1/3.$$ $$\mathrm{OY:}\ 0 = \mathrm{m}_2\mathrm{v}_2 - \mathrm{m}_3\mathrm{v}_{\mathrm{y}3};\ \mathrm{mv}_2 = 3\mathrm{mv}_{\mathrm{y}3};\ \mathrm{v}_{\mathrm{y}3} = \mathrm{v}_2/3.$$ $$\mathrm{v}_3 = \sqrt{\mathrm{v}_{x3}^2 + \mathrm{v}_{y3}^2} = \sqrt{10^2 + 10^2} \frac{\mathrm{m}}{\mathrm{s}} = \sqrt{200}\ \mathrm{m/s} \approx 14.14\ \mathrm{m/s}.$$

Answer: $\mathrm{v}_3 = 14.14\ \mathrm{m/s}$

http://www.AssignmentExpert.com/