A Two-Body Collision with a Spring A block of mass m 1 = 1.50 kg initially movin

Question

A Two-Body Collision with a Spring

A block of mass m1 = 1.50 kg initially moving to the right with a speed of 5.00 m/s on a frictionless, horizontal track collides with a light spring attached to a second block of mass m2 = 6.50 kg initially moving to the left with a speed of 2.50 m/s as shown in figure (a). The spring constant is 550 N/m.

What if m1 is initially moving at 3.2 m/s while m2 is initially at rest?

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

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

(1) Initially the block have kinetic energy.
K.E. = (1/2)(m)(v^2)
K.E. = (1/2)(4.7 kg)(2.6 m/s)^2 = 15.886 J.
When the box of running shoes collided with the 2.4-kg , then by conservation of momentum...
(Mr + Mb)before = (Mr + Mb)after
m1v1 + m2v2 = (m1 + m2)v
(4.7kg)(2.6 m/s) + (2.4 kg)(0) = (4.7kg + 2.4kg)(v)
v = [(4.7 kg)(2.6 m/s)]/(4.7 kg + 2.4 kg)
v = 1.72 m/s
Therefore, the two blocks left edge of the table with the HORIZONTAL speed of v = 1.72 m/s. Their K.E. due to this
is . . . K.E. = (1/2)(4.7 + 2.4)(1.72)^2 = 10.50 J

In that case the spring constant= KE of the Block

kx= 10.50

Now, the two boxes have P.E.g = (m1+m2)(g)(h)
P.E.g = (4.7 + 2.4)(9.8)(0.71m) = 49.40 J

By conservation of mechanical energy then, their kenetic energy is equal to: 10.50 J + 49.40 J = 59.9 J answer

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