Question

An 69-kg jogger is heading due east at a speed of 1.6 m/s. A 63-kg jogger is heading 14 ° north of east at a speed of 1.5 m/s. Find (a) the magnitude and (b) the direction of the sum of the momenta of the two joggers. Describe the direction as an angle with respect to due east.

Answer 1

Answer:

$P=203.3819375kg.m/s\\P\approx203.4kg.m/s$

$\theta=6.4541\textdegree North East\\\theta\approx6.5\textdegree North East$

Explanation:

From the question we are told that

Mass of first jogger $M=69kg$

Speed$v_1=1.6m/s$

Direction $d= East$

Mass of 2nd jogger $M=63kg$

Speed $v_2=1.5m/s$

Direction $d= 14\textdegree north$

Generally equation for momentum along the the horizontal is mathematically given as

$P_x=m_1v_1 +m_2v_2cos\theta$

$P_x=(69)*(1.6) +(63)*(1.5)cos14$

$P_x=202.0929461kg.m/s$

Generally equation for momentum along the the vertical is mathematically given as

$P_y=m_2v_2cos\theta$

$P_x=22.86161913kgm/s$

a)Generally the magnitude of momentum is mathematically given by

$P=\sqrt{(P_x)^2+(P_y)^2}$

$P=\sqrt{(202.0929461)^2+(22.86161913)^2}$

$P=\sqrt{41364.21249}$

$P=203.3819375kg.m/s$

$P\approx203.4kg.m/s$

b) Generally the angle $\theta$ is mathematically given by

$\theta=tan^-^1\frac{py}{px}$

$\theta=tan^-^1\frac{22.86161913}{202.0929461}$

$\theta=tan^-^1(0.1131242805)$

$\theta=6.4541\textdegree North East$