Question

1. a 0.40 kg model airplane is travelling 20 km/h toward the south. a 0.50 kg model airplane, travelling 25 km/h in a direction 20° east of south, collides with the first model airplane. the two planes stick together on impact. what is the direction and magnitude of the velocity of the combined wreckage immediately after the collision? 7. what impulse is needed to change the velocity of a 90 kg football - player

Explanation:

Step1: Resolve initial velocities into components

For the first plane with mass $m_1 = 0.40\ kg$ and velocity $v_1=20\ km/h$ south - ward:
The velocity vector $\vec{v}_{1}=- 20\hat{j}\ km/h$.
For the second plane with mass $m_2 = 0.50\ kg$ and velocity $v_2 = 25\ km/h$ at $20^{\circ}$ east of south.
The $x$ - component of the second - plane's velocity is $v_{2x}=25\sin20^{\circ}\hat{i}\ km/h$ and the $y$ - component is $v_{2y}=-25\cos20^{\circ}\hat{j}\ km/h$.

Step2: Use the law of conservation of momentum

The total initial momentum in the $x$ - direction is $p_{ix}=m_2v_{2x}=0.50\times25\sin20^{\circ}\ kg\cdot km/h$.
$p_{ix}=0.50\times25\times0.342 = 4.275\ kg\cdot km/h$.
The total initial momentum in the $y$ - direction is $p_{iy}=m_1v_{1}+m_2v_{2y}=0.40\times(- 20)+0.50\times(-25\cos20^{\circ})\ kg\cdot km/h$.
$p_{iy}=-8 - 0.50\times25\times0.9397=-8 - 11.74625=-19.74625\ kg\cdot km/h$.
The total mass after the collision is $m = m_1 + m_2=0.40 + 0.50=0.90\ kg$.
The $x$ - component of the final velocity is $v_{fx}=\frac{p_{ix}}{m}=\frac{4.275}{0.90}\ km/h\approx4.75\ km/h$.
The $y$ - component of the final velocity is $v_{fy}=\frac{p_{iy}}{m}=\frac{-19.74625}{0.90}\ km/h\approx - 21.94\ km/h$.

Step3: Calculate the magnitude of the final velocity

The magnitude of the final velocity $v_f=\sqrt{v_{fx}^{2}+v_{fy}^{2}}$.
$v_f=\sqrt{(4.75)^{2}+(-21.94)^{2}}=\sqrt{22.5625 + 481.3636}=\sqrt{503.9261}\approx22.45\ km/h$.

Step4: Calculate the direction of the final velocity

The direction $\theta$ is given by $\tan\theta=\frac{v_{fx}}{|v_{fy}|}$.
$\tan\theta=\frac{4.75}{21.94}\approx0.2165$.
$\theta=\arctan(0.2165)\approx12.1^{\circ}$ east of south.

Answer:

The magnitude of the velocity of the combined wreckage is approximately $22.45\ km/h$ and the direction is approximately $12.1^{\circ}$ east of south.