Question

A particle of mass 500 g is at rest. It is free to move along a straight line. The power delivered to the particle varies with time according to the following graph :
The momentum of the particle at t=5s is

• A. 2√5 Ns
• B. 5√2 Ns
• C. 5 Ns
• D. 5.5 Ns

Answer 1

The Correct Option is C

Given:
Mass of the particle \( m = 500 \, \text{g} = 0.5 \, \text{kg} \)
The particle starts from rest, so initial velocity \( u = 0 \)
The power delivered to the particle is given as a function of time according to the graph.
Step 1: Relationship between power and work done The power delivered to the particle is related to the rate of work done on it. The work done at any instant is the integral of power over time: \[ W = \int P \, dt \] Since the power-time graph is a piecewise function, we need to calculate the area under the graph between 0 and 5 seconds to find the work done.
Step 2: Total work done from 0 to 5 seconds From the graph, we can see that the power increases linearly from 0 to 10 watts between \( t = 0 \) and \( t = 5 \) seconds. The area under the power-time curve is the area of a triangle, given by: \[ W = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \, \text{s} \times 10 \, \text{W} = 25 \, \text{J} \]
Step 3: Using work-energy theorem The work done on the particle is equal to the change in its kinetic energy. The kinetic energy at any time \( t \) is: \[ KE = \frac{1}{2} m v^2 \] The work done is: \[ W = \Delta KE = \frac{1}{2} m v^2 \] Substitute the known values: \[ 25 = \frac{1}{2} \times 0.5 \times v^2 \] \[ 25 = 0.25 v^2 \] \[ v^2 = \frac{25}{0.25} = 100 \] \[ v = 10 \, \text{m/s} \]
Step 4: Momentum at \( t = 5 \) seconds Momentum is given by: \[ p = mv \] Substitute the values for \( m \) and \( v \): \[ p = 0.5 \times 10 = 5 \, \text{Ns} \] Thus, the momentum of the particle at \( t = 5 \) seconds is \( 5 \, \text{Ns} \).

Therefore, the correct answer is (C) 5 Ns.