1)A 83.9 kg toy rocket is sent flying straight up into the air with an accelerat

Question

1)A 83.9 kg toy rocket is sent flying straight up into the air with an acceleration of 7.50 m/s2. After 4.20 s it explodes and breaks apart into two pieces. The heavier piece has a mass of 65.0 kg. If the lighter piece is able to reach a maximum height of 182 m above the ground, what is the speed of the heavier piece of the toy rocket just after the explosion?

2) Two identical ping pong balls are rolling toward each other. Ball one is shot to the east at a speed of 1.00 m/s and ball two is shot to the west at 4.20 m/s. They do not hit head on but have a glancing collision that deflects ball two by 90°. Ball two is now moving to the north at 1.46 m/s. What is the speed of ball one after the collision?

Explanation / Answer

1) after 4.20 sec,

h1 = v0 t + a t^2 /2 = 0 + (7.50)(4.20^2)/2 = 66.15 m

v1 = v0 + at = 31.5 m/s

for lighter piece kg:

vf^2 - v0^2 = 2 a d

0 ^2 - v2^2 = 2 (-9.81) (182 - 66.15)

v2 = 47.7 m/s speed just after explosion.

Applying momentum conservation,

(83.9) (31.5) = (65v) + ((83.9 - 65)(47.7) )

v = 26.8 m/s upward ......Ans

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