Question

The velocity of a body of mass $15kg$ increases from $5m{s}^{-1}$ to $10m{s}^{-1}$ when a force acts on it for $2s$. What is the gain in momentum per second?

Answer 1

Step 1: Given Data:

Mass, $m=15kg$

Initial velocity, $v_i=5m{s}^{-1}$

Final velocity, $v_f=10m{s}^{-1}$

Step 2: Formula used: Gain in momentum,$∆p=m\left(v_f-v_i\right)$, where $v_f$ and $v_i$ are final and initial velocity respectively.

Step 3: Calculation of gain in momentum:

$$\begin{gathered} ∆p=15\left(10-5\right) \\ ∆p=15\times 5 \\ ∆p=75kgm{s}^{-1} \end{gathered}$$

Gain in momentum for 2 s, $∆p=75kgm{s}^{-1}$

Gain in momentum per second,

$$\begin{gathered} \frac{∆p}{∆t}=\frac{75}{2}kgm{s}^{-1} \\ \frac{∆p}{∆t}=37.5kgm{s}^{-1} \end{gathered}$$