Question

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An object of mass 1 kg travelling in a straight line with the velocity of 10m/s collide with and sticks to a stationary wooden block of mass 5 kg. Then they both move together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

⛄Non- copied answer only⛄​

Answer 1

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Non copied Answer.....

Mass of the object, m1 = 1 kg

Velocity of the object before collision, v1 = 10 m/s

Mass of the stationary wooden block, m2 = 5 kg

Velocity of the wooden block before collision, v2 = 0 m/s

ˆ´ Total momentum before collision = m1 v1 + m2 v2

= 1 (10) + 5 (0) = 10 kg m sˆ’1

It is given that after collision, the object and the wooden block stick together.

Total mass of the combined system = m1 + m2

Velocity of the combined object = v

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

m1 v1 + m2 v2 = (m1 + m2) v

1 (10) + 5 (0) = (1 + 5) v

v = 10 / 6

= 5 / 3

The total momentum after collision is also 10 kg m/s.

Total momentum just before the impact = 10 kg m sˆ’1

Total momentum just after the impact = (m1 + m2) v = 6 — 5 / 3 = 10 kg ms-1

Hence, velocity of the combined object after collision = 5 / 3 ms-1...

Hope it helps you....

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Answer 2

Explanation:

QUESTION:-

An object of mass 1 kg travelling in a straight line with the velocity of 10m/s collide with and sticks to a stationary wooden block of mass 5 kg. Then they both move together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

ANSWER:-

GIVEN :-

Mass of the object, m1 = 1 kg

Velocity of the object before collision, v1 = 10 m/s

Mass of the stationary wooden block, m2 = 5 kg

Velocity of the wooden block before collision, v2 = 0 m/s

SOLUTION:-

ˆ´ Total momentum before collision = m1 v1 + m2 v2

= 1 (10) + 5 (0) = 10 kg m sˆ’1

It is given that after collision, the object and the wooden block stick together.

Total mass of the combined system = m1 + m2

Velocity of the combined object = v

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

m1 v1 + m2 v2 = (m1 + m2) v

1 (10) + 5 (0) = (1 + 5) v

v = 10 / 6

= 5 / 3

The total momentum after collision is also 10 kg m/s.

Total momentum just before the impact = 10 kg m sˆ’1

Total momentum just after the impact = (m1 + m2) v = 6 — 5 / 3 = 10 kg ms-1

Hence, velocity of the combined object after collision = 5 / 3 ms-1

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