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A 45 kg student is riding on a 7kg skateboard with a velocity of +4 m/s. The student jumps off the skateboard with a velocity of 1 m/s. Find the velocity of the skateboard after the student jumped off



Answer :

SuchandraC

According to conservation of momentum,

(m1+m2) v = m1v1 + m2v2

Thus, velocity of the skateboard, v2= 36m/s

How to calculate the velocity of skateboard?

As per the law of conservation of momentum  in an isolated system the total momentum of two or more bodies acting upon each other remains constant unless an external force is applied.

Thus,

(m1+m2) v = m1v1 + m2v2

v2= (m1+m2) v - m1v1 /m2

= (45+7) (4) -(45) (1) /7

= 253/7 = 36m/s

Hence, velocity of the skateboard, v2= 36m/s  

To know the law of conservation of momentum, https://brainly.com/question/7538238

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