Question

Mission MC2: Impulse and Momentum Change

A 6.0-kg object moving at 3.0 m/s encounters a 40-
Newton resistive force over a duration of 0.20
seconds. The final momentum of this object is ___kg• m/s
. m/s.
a. - 5.0
b. 0.444
c. 0.50
d. 2.22
e. 8.0
f. 10
g. 11
h. 15
i. 18
j. 26
k. 90
I. 200

Answer 1

8.0 I mean for 10, 10 is wrong

Step-by-step explanation:

Multiplying the force on the object by the time over which the force acts, multiplying the mass of the object by the velocity change of the object

Answer 2

`10\; {\rm kg \cdot m\cdot s^(-1)}`.

(Assume that there are no other unbalanced force on this object.)

Step-by-step explanation:

When an object of mass
`m` travels at a velocity of
`v`, the momentum
`p` of that object would be
`p = m\, v`.

In this question, the
`m = 6.0\; {\rm kg}` object was initially at a velocity of
`v = 3.0\; {\rm m\cdot s^(-1)}`. The initial momentum of this object would be:

`\begin{aligned} p &= m\, v \\ &= (6.0\; {\rm kg})\, (3.0\; {\rm m\cdot s^(-1)}) = 18\; {\rm kg\cdot m\cdot s^(-1)}\end{aligned}`

The question states that the external force is "resistive". In other words, this force opposes the motion of this object. Hence, while the magnitude of this force is
`40\; {\rm N}`, the vector value of this force would be
`(-40)\; {\rm N}`.

To find the impulse
`P` that a force
`F` has exerted, multiply the force vector by the duration
`\Delta t` over which this force was applied.

For example, the
`(-40)\; {\rm N}` force in this question was applied over a period of
`\Delta t = 0.20\; {\rm s}`. This force would have exerted an impulse
`J` of:

`\begin{aligned}J &= F\, \Delta t \\ &= (-40\; {\rm N})\, (0.20\; {\rm s}) = (-8.0)\; {\rm N\cdot s}\end{aligned}`.

Note that
`1\; {\rm N}` is equivalent to
`1\; {\rm kg\cdot m\cdot s^(-2)}`. Therefore, the unit of impulse
`{\rm N\cdot s}` would be equivalent to
`{\rm (kg\cdot m\cdot s^(-2))\cdot s}`, which simplifies to
`{\rm (kg\cdot m\cdot s^(-1))}`.

Thus, the impulse on this object
`(-8.0)\; {\rm N\cdot s}` would be equivalent to
`(-8.0)\; {\rm kg\cdot m\cdot s^(-1)}`.

The total impulse on an object is also equal to the change in the momentum of the object. Assuming that there are no other unbalanced force on this object, the total impulse on this object would be
`(-8.0)\; {\rm kg\cdot m\cdot s^(-1)}`. The momentum of this object would become:

`\begin{aligned}& 18.0\; {\rm kg \cdot m\cdot s^(-1) + (-8.0)\; {\rm kg\cdot m\cdot s^(-1)} \\ =\; & 10\; {\rm kg\cdot m\cdot s^(-1)} \end{aligned}`.