A block of mass m 1 = 1.2 kg initially moving to the right with a speed of 4.6 m

Question

A block of mass m1 = 1.2 kg initially moving to the right with a speed of 4.6 m/s on a frictionless, horizontal track collides with a spring attached to a second block of mass m2 = 2.8 kg initially moving to the left with a speed of 2.7 m/s as shown in figure (a). The spring constant is 505N/m.

What if mi is initially moving at 2.6 m/s while m2 is initially at rest?

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

(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.)

? = m/s to the left

? = m/s to the right

A Two-Body Collision with a Spring A block of mass m1.2 kg initially moving to the right with a speed of 4.6 m/s on a frictionless, horizontal track collides with a spring attached to a second block of mass m2 - 2.8 kg initially moving to the left with a speed of 2.7 m/s as shown in figure (a). The spring constant is 505 N/m lf A moving block collides with another moving block with a spring attached: (a) before the collision and (b) at one instant during the collision

Explanation / Answer

(A)maximum compression will be when they have same velocity.

Applying momentum conservation,

(1.2 x 4.6) + (2.8 x -2.7) = (1.2 + 2.8) v

v = - 0.51 m/s

Applying energy conservation,

(1.2)(4.6^2)/2 + (2.8)(2.7^2)/2 = (1.2 + 2.8)(0.51^2)/2 + 505x^2 /2

x = 0.298 m

(b) after they are separated their velocity direction will be reversed.

for elastic collision, v1 + v2 = 4.6 + 2.7 = 7.3

v2 = 7.3 - v1

momentum conservation,

4x -0.51 = - 1.2v1 + 2.8v2

- 2.04 = - 1.2v1 + 20.44 - 2.8v1

v1 = 5.62 m/s ......Ans

v2 = 1.68 m/s ......Ans

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