Answered

Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.


A woman of mass 49 kg jumps off the bow
of a 78 kg canoe that is intially at rest.
If her velocity is 4.5 m/s to the right, what
is the velocity of the canoe after she jumps?
Answer in units of m/s i.
x x

Sagot :

LammettHash

The woman-canoe system has zero initial momentum. When the woman jumps from the canoe, the total momenta of the woman and canoe is conserved, so that

0 = (49 kg) (4.5 m/s) + (78 kg) v

where v is the velocity of the canoe. Solving for v, we find

(78 kg) v = - (49 kg) (4.5 m/s)

v = - 49/78 (4.5 m/s)

v ≈ -2.8 m/s

which is to say, the boat is given a velocity of 2.8 m/s to the left.

We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.