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A 0.025 kg bullet enters a 2.35-kg watermelon with a speed of 217 m/s

and exits the opposite side with a speed of 109 m/s. If the melon was

originally at rest, then what speed will it have as the bullet leaves its

opposite side?

Respuesta :

samuelonum1

Answer:

1.15m/s

Step-by-step explanation:

Given data

m1=0.025 kg

m2= 2.35kg

u1= 217 m/s

v1= 109 m/s

u2= 0

v2=???

Applying the law of conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

substitute

0.025*217+2.35*0= 0.025*109+2.35*v2

5.425= 2.725+ 2.35v2

5.425-2.725=2.35v2

2.7=2.35v2

divide both sides by 2.35

v2= 2.7/2.35

v2=1.15 m/s

Hence the speed of the watermelon will be 1.15m/s

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