Question

An object of mass 1 kg travelling in a straight line with a velocity of 10 m/s collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off 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 1

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⏩ Here, m1= 1 kg, u1= 10 m/s, m2= 5 kg, u2= 0

Let v be the velocity of the combined object after collision.

Now,

→The total momentum just before the impact= m1u1 + m2u2

= 1 × 10 + 5 × 0 = 10 kg m/s. Ans.

→ The total momentum just after the impact= (m1 + m2) v

= (1 + 5 ) v= 6 v kg m/s.

Now, by conservation of momentum,

6v= 10

or

v= 10/6 = 1.67 m/s. Ans.

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Answer 2

Given :

• Mass of object (m1) = 1 kg
• Mass of block (m2) = 5 kg
• Initial elocity of object (u1) = 10 m/s
• Initial vlocity of block (u2) = 0 m/s

(Initial velocity of block is zero because block is stationary)

Now,

Momentum = mv

According to question,

Initial Momentum = m1u1 + m2u2

=> 1 × 10 + 5 × 0

=> 10 + 0

=> 10 kg m/s

According to law of conservation of momentum.

The total momentum of the system cannot be changed.

So,

Initial momentum = Final momentum

So, Final momentum also is 10 kg m/s.

Also, we have to calculate the combined velocity of the object.

Let combined velocity of object = v

So,

Final Momentum = m1v + m2v

=> 10 = (m1 + m2) v

=> 10 = (1 + 5)v

=> 10 = 6v

=> v = 1.67 m/s

∴ Total momentum just before the impact and just after the impact is 10 kg m/s.

Combined velocity of the object is 1.67 m/s.