Question

A block of mass m1 moving with speed v1 undergoes a completely inelastic collision with a stationary block of mass m2. The blocks then move, stuck together, at speed v2. After a short time, the two-block system collides inelastically with a third block, of mass m3, which is initially stationary. The three blocks then move, stuck together, with speed v3 (Figure 1) . All three blocks have nonzero mass. Assume that the blocks slide without friction.

Answer 1

a)

v2 = m1 . v1 / (m1 + m2)

v2 / v1 = m1 / ( m1 + m2)

b)

Kinetic energy of the original block

k1 = ½m1v1²

Velocity after the first two blocks join
use conservation of momentum

v2 = m1v1 / (m1 + m2)

kinetic energy after first collision

k2 = ½(m1 + m2)(m1v1 / (m1 + m2))²
k2 = ½(m1v1)² / (m1 + m2))

k2/k1 = ½(m1v1)² / (m1 + m2)) / ½m1v1²
k2/k1 = m1 / (m1 + m2)

c)

v3 = m1 . v1 / ( m1 + m2 + m3 )

v3 / v1 = m1 / ( m1 + m2 + m3 )

d)

velocity after second collision
Let
m1 + m2 = m11


v3 = m11v2 / (m11 + m3)
v3 = m11( m1v1 / m11) / (m11 + m3)
v3 = m1v1 / (m11 + m3)
v3 = m1v1 / (m1 + m2 + m3)

kinetic energy after second collision

k3 = ½(m11 + m3)v3²
k3 = ½(m11 + m3)(m1v1 / (m11 + m3))²
k3 = ½(m1²v1² / (m11 + m3))
k3 = ½(m1²v1² / (m1 + m2 + m3))

k3 / k1 = ½(m1²v1² / (m1 + m2 + m3)) / ½m1v1²
k3 / k1 = m1 / (m1 + m2 + m3)

e)

k4/k1 = m1 / (m1 + m2 + m3 + m4)