Block 1, of mass m1 = 6.70 kg , moves along a frictionless air track with speed

Question

Block 1, of mass m1 = 6.70 kg , moves along a frictionless air track with speed v1 = 31.0 m/s . It collides with block 2, of mass m2 = 47.0 kg , which was initially at rest. The blocks stick together after the collision. (Figure 1)

Part A

Find the magnitude pi of the total initial momentum of the two-block system.

Express your answer numerically.

Part B

Find vf, the magnitude of the final velocity of the two-block system.

Express your answer numerically.

Part C

What is the change K=KfinalKinitial in the two-block system's kinetic energy due to the collision?

Express your answer numerically in joules.

Explanation / Answer

part A

initial momentum Pi = m1*v1 + m2*v2

Pi = (6.7*31)+(47*0) = 207.7 kg m/s

part B)

final momentum Pf = (m1+m2)*vf

from moementum conservation

Pi = Pf

vf = pi/(m1+m2)

vf = 3.86 m/s

partc)

Kinitial = 0.5*m1*v1^2 = 0.5*6.7*31^2 = 3219.35 J

Kfinal = 0.5*(m1+m2)*vf^2 = 0.5*(6.7+47)*3.86^2 = 400.05426

dK = Kfinal - Kinitial

dK = 400.05426 - 3219.35

dK = -2819.29574 J

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