A block of mass m1 = 1.4 kg initially moving to the right with a speed of 3.7 m/

Question

A block of mass m1 = 1.4 kg initially moving to the right with a speed of 3.7 m/s on a frictionless, horizontal track collides with a spring attached to a second block of mass m2 = 4.3 kg initially moving to the left with a speed of 2.6 m/s as shown in figure (a). The spring constant is 511N/m.

What if m1 is initially moving at 2.6 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

I am completely lost and can not get the answer. If someone could help me out it would be greatly appreciated! Thanks!

Explanation / Answer

a) Applying the conservation of momentum,
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'
m1 * (v1 - v1') + m2 * v2 = m2 * v2'
v2' = m1/m2 * (v1 - v1') + v2
Let the direction to the right be +, the direction to the left be -.
So v1 = 4, v2 = -2.5, v1' = 3
Therefore,
v2' = 1.6/2.1 (4-3) - 2.5 = -1.74
Therefore, block2 is moving to the left with a velocity of 1.74 m/s
b) Applying the conservation of energy, some of the energy after the collision must have transferred to the spring.
The energy stored in the spring is E = 1/2 * kx^2
Here, k is the spring constant.
1/2 * m1 * v1^2 + 1/2 * m2 * v2^2 - 1/2 * m1 * v1'^2 - 1/2 * m2 * v2'^2 = 1/2 * kx^2
m1 (v1^2 - v1'^2) + m2 (v2^2 - v2'^2) = kx^2
1.6 * (4^2 - 3^2) + 2.1 (2.5^2 - 1.74^2) = 11.2 + 3.22 = 600 * x^2
14.42 / 600 = x^2
x = 0.16 m

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