Question

In: Physics

A block of mass m1 = 2.9 kg initially moving to the right with a speed...

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

Solutions

Expert Solution

(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


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