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प्रश्न
Figure 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 ?
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उत्तर १
A ball rebounding between two walls located between at x = 0 and x = 2 cm; after every 2 s, 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 2 s. 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)xx10^2)/(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
उत्तर २
This graph can be of a ball rebounding between two walls situated at position 0 cm and 2 cm. The ball is rebounding from one wall to another, time and again every 2 s with uniform velocity.
Impulse, Here velocity = `("displacement")/"time" = 2/(100xx2) = 0.01 ms^(-2)`
Initial momentum = mu = `0.04 xx 0.01 = 4 xx 10^(-4) "kg ms"^(-1)`
Final momentum = mv = `0.04 x (-0.01) = -4 xx 10^(-4) "kg ms"^(-1)`
Magnitude of impulse = Change in momentum
=`(4xx 10^(-4)) -(-4xx10^(-4)) = 8 xx 10^(-4) kg ms^(-1)`
Time between two consecutive impulses is 2 s.i.e the ball receive an impulse every 2 s
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