Question

solvethis question ​

Answer 1

Answer:

Hey mate here is your answer:

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→Mass of bullet= 10g= 0.01kg

→Velocity of bullet= 1.5m/s

→Mass of plank= 90g=0.09kg

→Total momentum before collision= (mass of plank) x (velocity)

→= (0.01x1.5) + (0.09x0)=0.015 kg m/s

→Total momentum after collision= (mass of bullet +mass of plank) x (velocity)

=(0.1 kg x V) =0.1 V kg m/s

According to law of conservation of momentum,

→Total momentum before collision=Total momentum after collision.

$m_1v_1=m_2v_2$.

→0.015kg m/s =0.1 v kg m/s

Thus, velocity acquired by the plank and the bullet = 0.15m/s.

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Hope it helps you.

Mahapatra,singhania or chakraborty ki jaiho!

Answer 2

Answer:

→Mass of bullet= 10g= 0.01kg

→Velocity of bullet= 1.5m/s

→Mass of plank= 90g=0.09kg

→Total momentum before collision= (mass of plank) x (velocity)

→= (0.01x1.5) + (0.09x0)=0.015 kg m/s

→Total momentum after collision= (mass of bullet +mass of plank) x (velocity)

=(0.1 kg x V) =0.1 V kg m/s

According to law of conservation of momentum,

→Total momentum before collision=Total momentum after collision.

m_1v_1=m_2v_2m

1

v

1

=m

2

v

2

.

→0.015kg m/s =0.1 v kg m/s

Thus, velocity acquired by the plank and the bullet = 0.15m/s.

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Hope it helps you.

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