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**

**===============================**

**B)**

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

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

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