Question

momentum, collisions problem - solving
two novice skaters are on the ice at the same time. skater 1,
who weighs 50 kg, is travelling east at 3.0 m/s and collides
with skater 2, who weighs 60 kg and is travelling north at 4.0
m/s. they grab each other as they collide and cling together
to avoid falling. what is their resulting velocity after the
collision as they strive to stay upright?

Explanation:

Step1: Calculate the initial momentum in x - direction

The initial momentum of skater 1 in the x - direction (\(p_{x1}\)) is given by \(p = mv\). For skater 1, \(m_1 = 50\space kg\) and \(v_{x1}=3.0\space m/s\). So \(p_{x1}=m_1v_{x1}=50\times3.0 = 150\space kg\cdot m/s\). Skater 2 has no initial momentum in the x - direction (\(p_{x2} = 0\)). The total initial momentum in the x - direction \(p_{x}=p_{x1}+p_{x2}=150\space kg\cdot m/s\). After the collision, the combined mass \(M=m_1 + m_2=50 + 60=110\space kg\). Let the x - component of the final velocity be \(v_x\). Using the law of conservation of momentum \(p_{x}=Mv_x\), so \(v_x=\frac{p_{x}}{M}=\frac{150}{110}\approx1.36\space m/s\)

Step2: Calculate the initial momentum in y - direction

The initial momentum of skater 2 in the y - direction (\(p_{y2}\)) is \(p = mv\). For skater 2, \(m_2 = 60\space kg\) and \(v_{y2}=4.0\space m/s\). So \(p_{y2}=m_2v_{y2}=60\times4.0 = 240\space kg\cdot m/s\). Skater 1 has no initial momentum in the y - direction (\(p_{y1} = 0\)). The total initial momentum in the y - direction \(p_{y}=p_{y1}+p_{y2}=240\space kg\cdot m/s\). Using the law of conservation of momentum \(p_{y}=Mv_y\), so \(v_y=\frac{p_{y}}{M}=\frac{240}{110}\approx2.18\space m/s\)

Step3: Calculate the magnitude of the final velocity

The magnitude of the final velocity \(v\) is given by the Pythagorean theorem \(v=\sqrt{v_x^{2}+v_y^{2}}\). Substituting \(v_x\approx1.36\space m/s\) and \(v_y\approx2.18\space m/s\), we get \(v=\sqrt{(1.36)^{2}+(2.18)^{2}}=\sqrt{1.85 + 4.75}=\sqrt{6.6}\approx 2.57\space m/s\)

Step4: Calculate the direction of the final velocity

The direction \(\theta\) (with respect to the x - axis) is given by \(\tan\theta=\frac{v_y}{v_x}\). \(\tan\theta=\frac{2.18}{1.36}\approx1.6\). So \(\theta=\arctan(1.6)\approx58^{\circ}\) north of east

Answer:

Their resulting velocity is approximately \(2.6\space m/s\) at an angle of \(58^{\circ}\) north of east.