# Impulse: Definition and Application: weight of a person in lift | Physics Grade XI

### impulse # Definition and Application of Impulse

Impulse

Impulse is the amount of force applied to a body for a very small amount of time. An impulsive force changes with time i.e. it is not constant. It is measured by taking the product of the average force applied and the time interval for which it is applied. i.e. impulse = Favg . *  delta t

According to Newton’s law of motion,

F = dp/dt

Or, F. dt = dp

Integrating both sides with limit, we get,

tʃ0 F. dt = p2ʃp1 dp = p2 – p1

So, impulse = p2 – p1

Application of impulse

1. While catching a cricket ball, the player lowers his hand to increase the time of impact. As F * t = change in momentum, greater the time of impact smaller is the force and vice-versa. However, the change in momentum or impulse is same in both cases
2. A person falling from a height into the cemented floor gets more injury than falling on the sand, because falling into cemented floor, the person comes in rest in short time but a person landing on sand takes more time to come in rest which decreases the impulse and hence gets less injury.

Weight of a person in a lift:

(i) When lift is at rest

If a man of mass 'm' standing on a lift at rest, mg works downwards whereas the reaction ‘R’ takes place just opposite to mg.

Since, it is at rest, acceleration and net force acting is zero, so we can write,

R – mg =0

R = mg

Therefore, the apparent weight is equal to the actual weight of the man

(ii) When lift is moving with uniform velocity in any direction (upward or downward)

If a man of mass ‘m’ standing on a lift moving with constant speed, mg works downwards whereas the reaction ‘R’ takes place just opposite to mg. As it is moving uniformly acceleration = 0, and Net force =0,

R – mg =0

R = mg

Therefore, the apparent weight is equal to

the actual weight of the man.

(iii) When lift is accelerating upwards If a man of mass ‘m’ standing on a lift moving upward with acceleration a, mg works downwards whereas the reaction ‘R’ works in an upward direction. Then,

F = R -mg

Or, ma = R -mg

Or, R = mg + ma

Or, R = m (g+a)

So, the apparent weight is more than the actual weight of the man.

(iv) When lift is accelerating downwards If a man of mass ‘m’ standing on a lift moving downward with acceleration a, mg works downwards whereas the reaction ‘R’ works in the upward direction. Then,

F = mg -R

Or, ma = mg -R

Or, R = mg - ma

Or, R = m (g-a)

So, the apparent weight (reaction of floor) is less than the actual weight of the man.

(v) While Free fall

If a man of mass ‘m’ standing on a lift moving downward with acceleration a, mg works downwards whereas the reaction ‘R’ works in the upward direction. Then, a = g

F = mg -R

Or, ma = mg -R

Or, R = mg - ma

Or, R = m (g-a)

Or, R = m(g-g)

Or, R = 0

Therefore, a person on the lift experiences weightlessness.

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