Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?
Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?
A ball rebounding between two walls located between at x = 0 and x = 2 cm; after every 2s, the ball receives an impulse of magnitude 0.08×10−2 kg m/s from the walls
The given graph shows that a body changes its direction of motion after every 2s. Physically, this situation can be visualized as a ball rebounding to and fro between two stationary walls situated between positions x = 0 and x = 2 cm. Since the slope of the x-t graph reverses after every 2 s, the ball collides with a wall after every 2 s. Therefore, ball receives an impulse after every 2 s.
Mass of the ball, m = 0.04 kg
The slope of the graph gives the velocity of the ball. Using the graph, we can calculate initial velocity (u) as:
u=(2−0)×10−2(2−0)=10−2 m/s
Velocity of the ball before collision, u=10−2 m/s
Velocity of the ball after collision, v=−10−2 m/s
(Here, the negative sign arises as the ball reverses its direction of motion.)
Magnitude of impulse = Change in momentum
=|mv−mu|=|0.04(v−u)|=|0.04(−10−2−10−2)=0.08×10−2 kg m/s
A ball rebounding between two walls located between at x = 0 and x = 2 cm; after every 2s, the ball receives an impulse of magnitude 0.08×10−2 kg m/s from the walls
The given graph shows that a body changes its direction of motion after every 2s. Physically, this situation can be visualized as a ball rebounding to and fro between two stationary walls situated between positions x = 0 and x = 2 cm. Since the slope of the x-t graph reverses after every 2 s, the ball collides with a wall after every 2 s. Therefore, ball receives an impulse after every 2 s.
Mass of the ball, m = 0.04 kg
The slope of the graph gives the velocity of the ball. Using the graph, we can calculate initial velocity (u) as:
u=(2−0)×10−2(2−0)=10−2 m/s
Velocity of the ball before collision, u=10−2 m/s
Velocity of the ball after collision, v=−10−2 m/s
(Here, the negative sign arises as the ball reverses its direction of motion.)
Magnitude of impulse = Change in momentum
=|mv−mu|=|0.04(v−u)|=|0.04(−10−2−10−2)=0.08×10−2 kg m/s
