Complete Question:
A block of mass m= 1/4 M, slides (from rest) down a 30º incline of a height h. At the bottom, it strikes a block of mass M, which is at rest on the horizontal surface. Assumins an elastic collision, and neglecting friction, determine:
What is the upper limit on the mass m if it is to rebound from m slide up the incline, stop and down slide the incline, and collide with M again?
Answer:
m ≤ 1/3 M
Explanation:
Assuming no external forces acting during the collision, momentum must be conserved, as follows:
m*v₀ = m*v₁ + M*v₂ (1)
where v₀ is the speed of m when at the bottom of the slide, v₁ is the speed of m after the collision, and v₂ is the speed of the mass M after the collision.
Applying energy conservation principle, we can find v₀ from the following equation:
1/2*m*v₀² = m*g*h⇒ v₀ = √2*g*h
Assuming that the collision is elastic, this means that the kinetic energy must be conserved, as follows:
1/2*m*v₀² = 1/2*m*v₁² + 1/2*M*v₂² (2)
Combining (1) and (2), we can find the values of v₁ and v₂, as follows:
v₁ = v₀* ((m-M)/(m+M)), v₂ = v₀*((2m/(m+M))
For mass m be able to rebound up the slide and collide again with the mass M after sliding down, the speed v₁ must be larger than v₂, so the following condition must be met:
(m-M) / 2m (in absolute value) ≥ 1
⇒M ≥ 3*m ⇒ m ≤ M/3 = m≤ 1/3 M
