Question

A) A 1500 kg weather rocket accelerates upward at 10m/s2. It explodes 1.7 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment traveled straight up and reached a maximum height of 520 m .What was the speed of the heavier fragment just after the explosion?

B) A tennis player swings her 1000 g racket with a speed of 9.0 m/s . She hits a 60 g tennis ball that was approaching her at a speed of 17 m/s . The ball rebounds at 40 m/s .If the tennis ball and racket are in contact for 11 ms , what is the average force that the racket exerts on the ball?

Answer 1

A)

positon of rocket after t = 1.7 s

yo = voy*t + (1/2)*ay*t^2

yo = (1/2)*10*1.7^2 = 14.45 m

speed of rocket after time t = 1.7 s

vo = a*t = 10*1.7 = 17 m/s ( upwards)

after breaking in to two fragments

m2 = 2*m1

for m1

at maimum poiint final velcoity v1' = 0

final position y = 530 m

v1'^2 - v1^2 = 2*a*(y - yo)

0 - v1^2 = -2*9.8*(530-14.45)

v1 = 100.52 m/s

from momentum conservation

toal momentum remains same

(m1+m2)*v = m1*v1 + m2*v2

(m1 + 2*m1)*17 = m1*100.52 + 2*m1*v2

3*17 = 100.52 + 2*v2

v2 = -24.76 m/s

the speed of heavier mass is 24.76 m/s downwards <<-----answer

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B)

average force F = m*(vf - vi)/t

F = 60*10^-3*(40 - (-17))/0.011 = 311 N <<-----answer