Question

A particle is projected with a velocity u making an angle theta with the horizaontal.what is the change in velocity when it is at the highest point

Answer 1

We know that at the highest point of its motion a projectile has only its horizontal component of velocity i.e Ucosθ

So,after breaking,one part can retrace its pathway if it will have the same velocity after the collsion in the opposite direction.

So,applying law of conservation of momentum,

Initial momentum was mUcosθ

After the collsion momentum became, −m2Ucosθ+m2v (where,v is the velocity of the other part)

So,equating we get,

mUcosθ=−m2Ucosθ+m2v

or, v=3Ucosθ

Answer 2

Answer:

We know that at the highest point of its motion a projectile has only its horizontal component of velocity i.e Ucosθ

So,after breaking,one part can retrace its pathway if it will have the same velocity after the collsion in the opposite direction.

So,applying law of conservation of momentum,

Initial momentum was mUcosθ

After the collsion momentum became, −m2Ucosθ+m2v (where,v is the velocity of the other part)

So,equating we get,

mUcosθ=−m2Ucosθ+m2v

or, v=3Ucosθ

Explanation: