Question

A girl of mass 50 kg jumps out of a rowing boat of mass 300 kg on to the bank with a horizontal velocity of 3 m/s3 m/s . With what velocity does the boat begin to move backwards​

Answer 1

Answer:

• Boat will move with a velocity 0.5 m/s backward.

Explanation:

Given

• Mass of girl, m = 50 kg
• Mass of boat, M = 300 kg

Initially, Boat would be at rest and Girl is moving with velocity 3 m/s, so

• Initial velocity of girl, u = 3 m/s
• Initial velocity of Boat, U = 0 m/s

After the jump of girl, Boat will start moving backwards with a certain velocity and Girl will standstill on the boat, so

• The Final velocity of girl, v = 0 m/s
• and Let, the final velocity of Boat = V

To find

• The Final velocity of Boat, V =?

Knowledge required

• Law of conservation of linear momentum

According to the law of conservation of linear momentum, the total momentum of the system before the collision is equal to the total momentum of the system after the collision.

• For two bodies of masses m and M, having initial velocity u and U and final velocities v and V. By the law of conservation of linear momentum

        m u + M U = m v + M V

Solution

Using law of conservation of linear momentum

→ m u + M U = m v + M V

[ putting known values ]

→ 50 × 3 + 300 × 0 = 50 × 0 + 300 × V

→ 150 + 0 = 0 + 300 V

→ 150 = 300 V

→ V = 150 / 300

→ V = 0.5  m/s

Therefore,

• Boat will move with a velocity of 0.5 m/s backward when the girl will jump on it.

Answer 2

Answer:

Given:-

${ \sf{ m_{1} = 50kg }}$

${ \sf{m_{2} = 300kg}}$

${ \sf{ v_{1} = 3m/s }}$

Find:-

what velocity does the boat begin to move backwards?

Solution:-

By law of conservation of momentum :-

${ \boxed{ \sf{ m_{1} v_{1} + m_{2} v_{2} = 0}}}$

${ \sf{50 \times 3 + 300 \times v_{2} = 0}}$

${ \sf{150 + 300 v_{2} = 0}}$

${ \sf{300 v_{2} = - 150 }}$

${ \sf{ v_{2} = \frac{ - 150}{300} }}$

${ \sf{ v_{2} = \frac{ - 1}{2} }}$

${ \sf{ v_{2} = 0.5}}$

Therefore, 0.5 m/s is the velocity of boat when it begins to move backwards.