Question

A 40.0-kg boy, riding a 2.50-kg skateboard at a velocity of +5.30 m/s across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the sidewalk is 6.00 m/s, 9.50° above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the skateboard's velocity relative to the sidewalk at this instant? Be sure to include the correct algebraic sign with your answer.

Answer 1

Answer:

$v_2 = -4.584\ m/s$

Explanation:

given,

mass of boy(m₁) = 40 kg

mass of the skateboard(m₂) = 2.50 kg

velocity of skateboard(V) = 5.30 m/s

speed of the boy(v₁) = 6 m/s at 9.50° above the horizontal

friction force is not acting in x-direction hence no we can apply conservation of momentum in x- direction

as the boy jumps over sidewalk skateboard cannot go through the sidewalk,  there must be some force in the y direction  so, momentum is not conserved.

$P_{before}x = P_{after}x$

$(m_1+m_2)V= m_1v_1+ m_2v_2$

$(40+2.5)\times 5.30 = 40 \times v cos (\theta )+ 2.5v_2$

$225.25 = 40 \times 6\times cos (9.50^0 )+ 2.5v_2$

$225.25 = 236.71+ 2.5v_2$

$v_2 = \dfrac{225.25-236.71}{2.5}$

$v_2 = -4.584\ m/s$

negative sign represent velocity of the skateboard is in the opposite direction of boy.