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Question

Figure 5.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s ? (Consider one-dimensional motion only).

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Solution

(a)

Given: The mass of the particle is 4kg.

The position time graph of the particle is shown below:



For t<0:

The particle is at rest for t<0. Therefore, no net force acts on it during this interval.

For t>4s:

The particle has constant position at this time interval. Since the particle does not move, no net force acts on it during this interval.

For 0<t<4s:

It can be seen in the graph that the position-time graph has constant slope. This means that it has constant velocity. Therefore, the acceleration of the particle is zero and no net force acts on the particle.

(b)

Given: The mass of the particle is 70kg.

The impulse is given by the change in momentum.

I=mvmu =m( vu ) (1)

Where, I is the impulse, v is the final velocity and u is the initial velocity.

For t=0:

The initial velocity (before point O) is 0m/s .

The final velocity (after point O) is the slope of the line OA.

v= 3 4 m/s

By substituting the given values in equation (1), we get

I=( 4kg )( 3 4 m/s 0m/s ) =3 kgm/s

Thus, the impulse at t=0s is 3 kgm/s .

For t=4s:

The initial velocity (before point A) is the slope of the line OA.

u= 3 4 m/s

The final velocity (after point A) is 0m/s .

By substituting the given values in equation (1), we get

I=( 4kg )( 0m/s 3 4 m/s ) =3 kgm/s

Thus, the impulse at t=4s is 3 kgm/s .


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