help Object 1 (mass=5.2 kg) traveling in the +x direction at a speed of 10.94m/s

Question

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Object 1 (mass=5.2 kg) traveling in the +x direction at a speed of 10.94m/s undergoes an inelastic collision with Object 2 (mass=4.20kg) which is at rest. Object two flies off at a speed of 8.53m/s and an angle theta 2 = 45.6 degree. The diagram below is just a sketch and is not meant to have accurate angles. It is possible that Object 1 travels back and to the left, though it isnot possible for it to travel in the -y direction. (Do you know why?) What is the total momentum in x direction after the collision?

Explanation / Answer

M1 = 5.2 kg

M2 = 4.2 kg

V1x = 10.94 m/s'

V1y = 0

V2x = 0

V2y = 0

V2x' = 8.53*cos(45.6)

V2y' = 8.53*sin(45.6)

let, V1x' = V1' * cos(a)

       V2y' = V1' * sin(a)

by momentum conservation

in x direction,

M1*V1x + M2*V2x= M1*V1x' + M2*V2x'

5.2*10.94 + 4.2*0 = 5.2*V1'*cos(a) + 4.2*8.53*cos(45.6)

56.888 = 5.2*V1'*cos(a) + 25.066

5.2*V1'*cos(a) = 31.822

in y direction,

M1*V1y + M2*V2y = M1*V1y' + M2*V2y'

5.2*0 + 4.2*0 = 5.2*V1'*sin(a) + 4.2*8.53*sin(45.6)

0 = 5.2*V1'*sin(a) + 25.6

5.2*V1'*sin(a) = - 25.6

tan(a) = 0.805

a = 218.834 degree from + x axis

V1' = 31.822/[5.2*cos(218.834)] = - 7.8561 m/s

Total momentum in x direction after collision remain same as before collision which is equal to 56.888 kg.m/s

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