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 ?

#### Solution 1

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

#### Solution 2

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