A Two-Body Collision with a Spring A block of mass m1 = 2.4 kg initially moving

Question

A Two-Body Collision with a Spring A block of mass m1 = 2.4 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 = 4.1 kg initially moving to the left with a speed of 2.3 m/s as shown in figure (a The the collision and (b) at one instant during the collision spring constant is 548 N/m 2i /n /n A moving block collides with another moving block with a spring attached: (a) before (A) Find the velocities of the two blocks after the collision (B) During the collision, at the instant block 1 is moving to the right with a velocity of 0.6 m/s as in figure (b), determine the velocity of block 2 (C) Determine the distance the spring is compressed at that instant SOLVE IT (A) Find the velocities of the two blocks after the collision Conceptualize: With the help of figure (a), run an animation of the collision in your mind. Figure (b) shows an instant during the collision when the spring is compressed. Eventually, block 1 and the spring will again separate, so the system will look like figure (a) again but with different velocity vectors for the two blocks Categorize: Because the spring force is conservative, kinetic energy in the system is not transformed to internal energy during spring, we can categorize the collision as being elastic the com pression of the spring. Ignoring any sound made when the block hits the

Explanation / Answer

Initial E = KE = ½mv² = ½ * 2.4kg * (2.8m/s)² = 9.408 J
max spring compression where both velocities are the same: conserve momentum:
2.4kg * 2.8m/s = (2.4 + 4.1)kg * v v = 1.033 m/s
which means the combined KE = ½ * (2.4 + 4.1)kg * (1.033m/s)² = 3.47 J

The remaining energy went into the spring:
U = (9.408 - 3.47) J = 5.938 J = ½kx² = ½ * 548N/m * x²
x = 0.1472 m (a)

(b) Momentum is still conserved:
6.5kg * 1.033m/s = 2.4kg * u + 4.1kg * v
6.7145 - 2.4u = 4.1v v = (6.7145 - 2.4u) / 4.1
and v² = (6.7145 - 2.4u)² / 4.1²
Energy is also conserved here:
½ * 2.4kg * u² + ½ * 4.1kg * v² = 9.408 J Plug in momentum result
1.2u² + 2.05(6.7145 - 2.4u)²/4.1² = 9.408
1.5426u² - 1.9167u - 6.728 = 0
u = -1.55 m/s, 2.8 m/s
Note that 2.8 m/s is the initial velocity, and can't be a solution here. A negative velocity is to be expected; so
m1's velocity u = -1.55 m/s
m2's velocity v = (6.7145 - 2.4*-1.55) / 4.1 = 2.545 m/s

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