Block 1, of mass m 1 = 7.70 kg , moves along a frictionless air track with speed
ID: 1600475 • Letter: B
Question
Block 1, of mass m1 = 7.70 kg , moves along a frictionless air track with speed v1 = 31.0 m/s . It collides with block 2, of mass m2 = 15.0 kg , which was initially at rest. The blocks stick together after the collision. (Figure 1)
1)Find the magnitude pi of the total initial momentum of the two-block system.
2)Find vf, the magnitude of the final velocity of the two-block system.
3)What is the change K=KfinalKinitial in the two-block system's kinetic energy due to the collision?
Before collision ms m1 After collisionExplanation / Answer
A) The total initial momentum is
m1v1 + m2v2 = pi
pi = 7.7 kg * 31.0 m/s + 15.0 kg * 0.0 since m2 was initially at rest, v2 = 0
pi = 238.7 kg.m/s <====
B) The mass of the coalesced body after collision is m1 + m2
vf is the final velocity of the two-body system
So, momentum is conserved, means:
m1v1 + m2v2 = (m1+m2)vf
or vf = (m1v1 + m2v2)/(m1+m2)
vf = pi / (m1+m2), pi computed above
vf = 238.7 kg.m/s /(7.7+15.0)kg
vf = 10.51 m/s <====== Ans
C) Change in KE
KE = Kf - Ki
KE = ½(m1+m2)vf² - ½m1v1² since v2 = 0 initially
KE = ½(7.7+15.0)10.51² - 0.5*7.7*31.0²
KE = - 2446.13 J
This is the loss of KE due to the collision