Question

PART A: find the direction of the total momentum for this system PART B: find the magnitude of the total momentum for this system

Answer 1

ANSWER

(A) 77.24° north of east

(B) 71.98 kg m/s

EXPLANATION

PART A and B

Let's draw this situation to understand it better,

First, we have to find the individual moments of the dog and the cat,

$p_{\text{dog}}=m_{\text{dog}}\cdot v_{\text{dog}}=26\operatorname{kg}\cdot2.7m/s=70.2\operatorname{kg}m/s$$p_{\text{cat}}=m_{\text{cat}}\cdot v_{\text{cat}}=5.3\operatorname{kg}\cdot3m/s=15.9\operatorname{kg}m/s$

The momentum of the system is the resultant vector of the two momentums. Geometrically,

Since the momentums of the dog and cat are in perpendicular directions, the momentum of the system is the hypotenuse of the right triangle where the momentums of the animals are the legs. Hence, we can find the angle alpha using the tangent of that angle,

$\tan \alpha=\frac{p_{\text{cat}}}{p_{\text{dog}}}$

Solving for alpha,

$\alpha=\tan ^{-1}\frac{p_{\text{cat}}}{p_{\text{dog}}}=\tan ^{-1}\frac{15.9}{70.2}\approx12.76\text{degrees}$

This angle and theta form a 90-degree angle, so they are complementary,

$\begin{gathered} \alpha+\theta=90 \\ \theta=90-\alpha=90-12.76=77.24 \end{gathered}$

So the direction of the total momentum of the system is 77.24 degrees north of east.

The magnitude can be found also using the properties of a right triangle. In this case, we can apply the Pythagorean theorem,

$p_{\text{sys}}=\sqrt[]{p^2_{\text{cat}}+p^2_{dog}}$$p_{\text{sys}}=\sqrt[]{15.9^2+70.2^2}\approx71.98\operatorname{kg}m/s$

The magnitude of the total momentum of the system is 71.98 kg m/s