Question

1. a hockey player of mass 82 kg is travelling north with a velocity of 4.1 m/s. he collides with a 76 kg player travelling east at 3.4 m/s. if the two players lock together momentarily, in what direction will they be going immediately after the collision? how fast will they be moving?

Explanation:

Step1: Calculate initial east - west momentum

The first player has mass $m_1 = 76$ kg and velocity $v_{1x}=3.4$ m/s (east - west direction). The second player has no initial velocity in the east - west direction. The initial east - west momentum $p_x$ is given by $p_x=m_1v_{1x}$. So, $p_x = 76\times3.4=258.4$ kg·m/s.

Step2: Calculate initial north - south momentum

The second player has mass $m_2 = 82$ kg and velocity $v_{2y}=4.1$ m/s (north - south direction). The first player has no initial velocity in the north - south direction. The initial north - south momentum $p_y$ is given by $p_y=m_2v_{2y}$. So, $p_y=82\times4.1 = 336.2$ kg·m/s.

Step3: Calculate the total final momentum

After the collision, the two players stick together, so the total mass $m = m_1 + m_2=76 + 82=158$ kg. Let the final velocity components be $v_x$ and $v_y$. By conservation of momentum, $p_x = mv_x$ and $p_y = mv_y$.

Step4: Calculate the magnitude of the final velocity

The magnitude of the final velocity $v$ is given by $v=\sqrt{v_x^2 + v_y^2}$. Since $v_x=\frac{p_x}{m}=\frac{258.4}{158}\approx1.635$ m/s and $v_y=\frac{p_y}{m}=\frac{336.2}{158}\approx2.128$ m/s. Then $v=\sqrt{(1.635)^2+(2.128)^2}=\sqrt{2.673 + 4.528}=\sqrt{7.201}\approx2.68$ m/s.

Step5: Calculate the direction of the final velocity

The direction $\theta$ (measured from the east - west axis) is given by $\tan\theta=\frac{v_y}{v_x}$. So, $\tan\theta=\frac{2.128}{1.635}\approx1.3$. Then $\theta=\arctan(1.3)\approx52.4^{\circ}$ north of east.

Answer:

The speed is approximately $2.68$ m/s and the direction is approximately $52.4^{\circ}$ north of east.