Question

A 3.0 kg particle is moving with a velocity

v(t)=(6.0 i +7.0t j) m/s.

(Assume momentum is in kg⋅m/s, force is in newtons, and t is in seconds. Express your answers in vector form. Use the following as necessary: t. Do not include units in your answers.) What is the momentum (as a function of time) of this particle?
p(t)=____kg⋅m/s

What is the net force acting on this particle?
F(t)=____ N

Answer 1

What is the momentum (as a function of time) of this particle?

p(t)= (18 i + 21t j) kg⋅m/s

The momentum is mass times velocity, so:

p = mv

p = (3.0 kg)(6.0 i + 7.0t j) m/s

= (18 i + 21t j) kg⋅m/s

What is the net force acting on this particle?

F(t) = (1/3 i - 7j) N

We can use Newton's Second Law, F = ma, to find the net force:

a = dv/dt

a = (0 i + 7 j) m/s2 (taking derivative of velocity)

F = ma

F = (3.0 kg) * (0 i + 7 j) m/s2

= (1/3 i - 7j) N

Answer 2

Momentum:
`p(t) = (18\, \vec{i} + 21\, t\, \vec{j})\; {\rm kg\cdot m\cdot s^(-1)}`.

Net force:
`F(t) = (0\, \vec{i} + 21\, \vec{j}) \; {\rm N}`.

Step-by-step explanation:

Velocity, momentum, and force are all vector quantities.

To find momentum, multiply velocity by the mass
`m` of the object. Given that
`m = 3.0\; {\rm kg}`:

`\begin{aligned} p(t) &= m\, v(t) \\ &= \left((3.0)\, (6.0 \, \vec{i}) + (3.0)\, (7.0\, t\, \vec{j})\right)\; {\rm kg \cdot m\cdot s^(-1)} \\ &= (18\, \vec{i} + 21\, t\, \vec{j})\; {\rm kg \cdot m\cdot s^(-1)} \end{aligned}`.

Net force can be found by multiplying the acceleration of the object by its mass. To find acceleration, take the first derivative of velocity:

`\begin{aligned} a(t) &= \left((d)/(dt)[6.0]\, \vec{i} + (d)/(dt)[7.0\, t]\, \vec{j}\right)\; {\rm m\cdot s^(-2)}\\ &= \left(0\, \vec{i} + 7.0\, \vec{j}\right)\; {\rm m\cdot s^(-2)}\end{aligned}`.

Multiply acceleration by the mass of the object to find the net force:

`\begin{aligned}F(t) &= m\, a(t) \\ &= \left((3.0)\, (0\, \vec{i}) + (3.0)\, (7.0\, \vec{j})\right)\; {\rm N} \\ &= (0\, \vec{i} + 21\, \vec{j})\; {\rm N}\end{aligned}`.