Question

A block of mass m1 = 2.9 kg initially moving to the right with a speed of 4.3 m/s on a frictionless, horizontal track collides with a spring attached to a second block of mass m2 = 5 kg initially moving to the left with a speed of 2.8 m/s as shown in figure (a). The spring constant is 572 N/m.

What if m1 is initially moving at 2.2 m/s while m2 is initially at rest?

(a) Find the maximum spring compression in this case. x = m

(b) What will be the individual velocities of the two masses (v₁ and v₂) after the spring extended fully again? (That is, when the two masses separate from each other after the collision is complete.)

v1	=	m/s to the left
v2	=	m/s to the right

Answer 1

(a) Initial momentum of the system = m₁ u₁ + m₂u₂
since mass 2 is at rest, therefore (u₂) = 0
= 2.9*2.2 = 6.38 kg-m/s------------(1)
Now consider after the collision mass 1 compresses the spring therefore both block will move simultaneously, therefore
Final momentum of the system = (m₁ +m₂)V = (2.9+5)V ---------(2)
where V is the velocity after the collision.
Now applying the conservation of momentum
(2.9+5)V = 6.38
V = 0.808 m/s
Now Initial energy of the system = (1/2)m₁u₁² = (1/2)*(2.9)*(2.2)² = 7.018 J-----------(1)
Final energy of the sytem = [(1/2)*(m₁ +m₂)*(V²) ]+(1/2)kx² = [(1/2)*(2.9+5)*0.808² ]+(1/2)kx²
= 2.579 +(1/2)kx² ------------(2)
Now by conservation of energy
2.579 +(1/2)*572x² = 7.018
x = 0.124 m
(b) Now as we know that the collision is elastic therfore
velocity of approach = Velocity of seperation
u₁ - u₂ = V₂ - V₁
where u₁ is initial velocity of m₁ = 2.2 m/s
u₂ is initial velocity of m₂ = 0
V₁ is final velocity of m₁
V₂ is final velocity of m₂
2.2 - 0 = V₂ - V₁
V₂ - V₁ = 2.2 -----------(1)
Now final momentum of the system
= m₁V₁ + m₂V₂ = 2.9V₁ + 5V₂
applying conservation of momentum
2.9V₁ + 5V₂ = 6.38 ------(2)
On solving 1 and 2
V₁ = -0.584 m/s
-ve sign shows that the assumed direction is wrong the direction of final velocity will be toward left.
V₁ = 0.584 m/s toward left
V₂ = 1.615 m/s toward right