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 4 kg. 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>4 s: 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<4 s: 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 70 kg. The impulse is given by the change in momentum. I=mv−mu =m( v−u ) (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 0 m/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=( 4 kg )( 3 4  m/s −0 m/s ) =3  kgm/s Thus, the impulse at t=0 s is 3  kgm/s . For t=4 s: 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 0 m/s . By substituting the given values in equation (1), we get I=( 4 kg )( 0 m/s − 3 4  m/s ) =−3  kgm/s Thus, the impulse at t=4 s is −3  kgm/s .

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