Question

A model rocket is constructed with a motor that can provide a total impulse of 37 N*s The mass of the rocket is 1.040 kg. What is the speed that this rocket achieves when launched from rest? Neglect the effects of gravity and air resistance.

A 70kg person stands on a scale in an elevator. What is the apparent weight when the elevator is  moving upward at a constant speed.  

Answer 1

First problem:

Solution:

Given,

the total impulse = FΔt =37Ns

the mass of the rocket, m = 1.040kg

As the rocket starts from rest, the initial speed of the rocket, v_i = 0

Let the final speed of the rocket be v_f.

We know from Newton's second law of motion, that the rate of change of momentum of a body is equal to the force acting on it.

There, force acting on the rocket, F = Δp/Δt

FΔt = Δp

So, the impulse is equal to the change in momentum.

We know the momentum formula is p = mv.

And the change in momentum(Δp) is the difference between the final momentum(mv_f) and initial momentum(mv_i)

Δp = mv_f – mv_i

As Δp = FΔt,

FΔt  = mv_f – mv_i   

37 = (1.040*v_f – 1.040*0)

37 = 1.040*v_f

  v_f = 37/1.040

v_f = 35.576 m/s

So, the final speed achieved by the rocket will be 35.576 m/s.

Second problem:

Solution:

Given,

the mass of the person, m = 70kg

So, the actual weight of the person = mg = 70*10 = 700N.

In an elevator, the apparent weight of the person standing in the elevator is simply the normal force exerted by the elevator ( or the scale) on the person.

The weight of the person acts downward on the scale and a normal force is applied by the scale floor on the person in upward direction.

If the elevator is at rest, the normal force on the person is equal to the person's weight on the scale (or elevator floor).

In this case, the scale reads actual weight of the person.

But when the elevator is moving upward or downward, the normal force exerted on the person is not equal to the person's actual weight i.e., it is more or less than the person's actual weight.

When elevator is accelerated upward with an acceleration a, then the normal force on the person will be equal to mg + ma.

So, the scale reads the same, which is the apparent weight of the person. So, the apparent weight in this case is more than the actual weight.

When the elevator is accelerated downward with an acceleration a, then the normal force on the person will be equal to

mg–ma.

So, the apparent weight in this case is less than the actual weight.

But, when the elevator is moving upward or downward with constant speed, the acceleration of the elevator is zero.

So, the normal force applied by the scale on the person will be equal to his actual weight(mg).

So, the apparent weight of the person standing in an elevator, which is moving upward with a constant speed is equal to the person's true weight.

So, the apparent weight of the person = 700N.

So, the scale reads either 700N (in weight terms) or 70kg (in mass terms).

Note: g value is taken as 10m/s² . If we take g = 9.8m/s², then the apparent weight would be 686N.