Block 1, of mass m 1 = 1.70 kg , moves along a frictionless air track with speed

Question

Block 1, of mass m1 = 1.70 kg , moves along a frictionless air track with speed v1 = 15.0 m/s . It collides with block 2, of mass m2 = 39.0 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.

View Available Hint(s)

nothing

Submit

Part B

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

Express your answer numerically.

View Available Hint(s)

nothing

Submit

Part C

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

Express your answer numerically in joules.

View Available Hint(s)

nothing

Block 1, of mass m1 = 1.70 kg , moves along a frictionless air track with speed v1 = 15.0 m/s . It collides with block 2, of mass m2 = 39.0 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.

View Available Hint(s)

pi =

nothing

  kg?m/s  

Submit

Part B

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

Express your answer numerically.

View Available Hint(s)

vf =

nothing

  m/s  

Submit

Part C

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

Express your answer numerically in joules.

View Available Hint(s)

?K =

nothing

J

Explanation / Answer

Given,

mass, m1 = 1.7 kg, m2 = 39 kg

velocity, v1 = 15 m/s,

1) m1*v1 = 1.7 x 15 = 25.5 kg-m/s

2) Using law of conservation of momentum,

m1*v1 = (m1 + m2)*V

V = m1*v1/(m1 + m2) = 0.627 m/s

3) Ki = 1/2*1.7*15^2 = 191.25 J

Kf = 1/2(m1 + m2)*0.627^2 = 8 J

So ?K = 8 - 191.25 = -183.25 J

Comment in case any doubt, please rate my answer....

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