Question

a 46.0 - kg boy, riding a 2.50 - kg skateboard at a velocity of 5.20 m/s across a level sidewalk, jumps forward to leap over a wall. just after leaving contact with the board, the boys velocity relative to the sidewalk is 6.00 m/s, 7.70° above the horizontal. ignore any friction between the skateboard and the sidewalk. what is the skateboards velocity relative to the sidewalk at this instant? be sure to include the correct algebraic sign with your answer.

Explanation:

Step1: Apply the law of conservation of momentum in the horizontal direction

The law of conservation of momentum states that \(p_{initial}=p_{final}\). The initial momentum of the boy - skateboard system is \((m_{boy} + m_{skateboard})v_0\), where \(m_{boy}=46.0\space kg\), \(m_{skateboard}=2.50\space kg\), and \(v_0 = 5.20\space m/s\). The final momentum is \(m_{boy}v_{1x}+m_{skateboard}v_2\), where \(v_{1x}=v_1\cos\theta\) (the horizontal component of the boy's velocity after jumping), \(v_1 = 6.00\space m/s\), and \(\theta = 7.70^{\circ}\).
So, \((m_{boy}+m_{skateboard})v_0=m_{boy}v_1\cos\theta + m_{skateboard}v_2\)

Step2: Solve for \(v_2\)

Rearrange the equation from Step 1 for \(v_2\):

$$
LATEXBLOCK0
$$

First, calculate \(\cos(7.70^{\circ})\approx0.991\)

$$
LATEXBLOCK1
$$

$$
LATEXBLOCK2
$$

Answer:

\(-8.53\space m/s\)