Question

A particle of mass ‘m’ is moving in time ‘t’ on a trajectory given by

`vecr  = 10alphat^2hati + 5beta(t - 5)hatj`

Where α and β are dimensional constants.

The angular momentum of the particle becomes the same as it was for t = 0 at time t = ______ seconds.

Options

• 20

• 30

• 40

• 10

Answer 1

A particle of mass ‘m’ is moving in time ‘t’ on a trajectory given by

`vecr  = 10alphat^2hati + 5beta(t - 5)hatj`

Where α and β are dimensional constants.

The angular momentum of the particle becomes the same as it was for t = 0 at time t = 10 seconds.

Explanation:

The angular momentum of the particle is given by:

`vecL = m(vecr xx vecv)`

When you calculate this using the given position vector:

`vecr = 10alphat^2hati + 5beta (t - 5)hatj and vecv = 20alphathati + 5betahatj`

you get:

L_z ​= m ⋅ 50αβt(10 − t)

This expression becomes zero at t = 0 and again at t = 10 seconds.

So, the angular momentum becomes the same as it was at t = 0 after 10 seconds.