A man of mass m 1 = 70.0 kg is skating at v 1 = 8.00 m/s behind his wife of mass

Question

A man of mass m1 = 70.0 kg is skating at v1 = 8.00 m/s behind his wife of mass m2 = 50.0 kg, who is skating at v2 = 4.00 m/s. Instead of passing her, he inadvertently collides with her. He grabs her around the waist, and they maintain their balance.

(d) Write the general equation for conservation of momentum in terms of m1, v1, m2, v2, and final velocity vf.

(e) Solve the momentum equation for vf. (Use the following as necessary: m1, v1, m2, v2. Do not substitute numerical values, use variables only.)

(f) Substitute values, obtaining the numerical value for vf, their speed after the collision.

Explanation / Answer

given m1 = 70 kg , v1 =8 m/s , mass m2 = 50 kg , v2 = 4 m/s

after collison they move with common velocity Vf

d)

from conservation of momentum

m1v1+m2v2 =(m1+m2)Vf

e)

from the above equation

Vf = m1v1+m2v2/(m1+m2)

f)

Vf,couple = Vf, man =Vf,wife = 6.333 m/s

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