NEXT Chapter 07, Problem 22 A 48.0-kg boy, riding a 3.00-kg skateboard board, th

Question

NEXT Chapter 07, Problem 22 A 48.0-kg boy, riding a 3.00-kg skateboard board, the boy's velocity relative to the sidewalk is 6.00 m/s, 7.900 above the horizontal. Ignore any f 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 i riction 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 Vo Units the tolerance is +/-3%

Explanation / Answer

Let us consider the conservation of momentum: (horizontally in the absence of external forces like friction)

momentum (mv) of the (boy + skateboard) before jump = momentom after the jump

51*(5.2) = 48*6cos7.9 + 3v

=> v = -6.6889 m/s (in the backward direction)

2)

(a) Use conservation of energy to get the pre-impact velocity:

v = sqrt(2gh) =sqrt(2 * 9.8m/s2 * 1.21m) =4.869 m/s

It is in the positive direction

(b) The collission is elastic head-on collision, we know that

the relative velocity of approach = relative velocity of separation, or

4.869 m/s = v - u

v =4.869 + u

where v, u are the post-collision velocities of the block, ball respectively

Conserve momentum: initial p = final p

1.7kg *4.869 m/s = 1.7kg * u + 2.3 kg * v

substitute for v from above

8.2773 = 1.7u + 2.3(4.869 + u) = 11.1987 + 4u

-2.9214 = 4u

u = -0.730 m/s Thi is the velocity of the ball

v = 4.869 + (-0.73) = 4.138 m/s velocity of the block

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