Question

A 50 kg mass is sitting on a frictionless surface. an unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3 m/s.

a) what is the initial momentum of the mass?
b) what is the final momentum of the mass?
c) what was the force acting on the mass?
d) what was the impulse acting on the mass?

Answer 1

Initial momentum:
`0\; {\rm kg \cdot m \cdot s^(-1)}`.

Final momentum:
`150\; {\rm kg \cdot m\cdot s^(-1)}`.

Force on the mass (assuming that the force is constant):
`75\; {\rm N}`.

Impulse on the mass:
`150\; {\rm kg \cdot m \cdot s^(-1)}`.

Step-by-step explanation:

The momentum
`p` of an object is equal to the product of mass
`m` and velocity (a vector)
`v`.

The initial momentum of this mass was
`0\; {\rm kg \cdot m \cdot s^(-1)}` since the velocity of this object was initially
`0\; {\rm m\cdot s^(-1)}`.

At
`v = 3\; {\rm m\cdot s^(-1)}`, the momentum of this mass (
`m = 50\; {\rm kg}`) would be:

`\begin{aligned}p &= m\, v \\ &= 50\; {\rm kg} * 3\; {\rm m\cdot s^(-1)} \\ &= 150\; {\rm kg \cdot m \cdot s^(-1)}\end{aligned}`.

Assume that the external force
`F` on this mass is constant. By Newton's Second Law of motion, the external force on this mass would be equal to the rate of change in the momentum of this mass.

Since the momentum of this mass increased by
`\Delta v = 150\; {\rm kg \cdot m \cdot s^(-1)}` in
`\Delta t = 2\; {\rm s}`, the external force on this mass would be:

`\begin{aligned}F &= (\Delta v)/(\Delta t) \\ &= \frac{150\; {\rm kg \cdot m \cdot s^(-1)}}{2\; {\rm s}} \\ &= 75\; {\rm kg \cdot m \cdot s^(-2)} \\ &= 75\; {\rm N}\end{aligned}`.

The impulse of an external force on an object is equal to the change in the momentum of that object. Since the change in momentum of this mass was
`\Delta v = 150\; {\rm kg \cdot m \cdot s^(-1)}`, the corresponding impulse would also be
`150\; {\rm kg \cdot m\cdot s^(-1)}`:

`\begin{aligned}J &= \Delta v \\ &= 150\; {\rm kg \cdot m\cdot s^(-1)} \\ &= 150\; {\rm N \cdot s}\end{aligned}`.