Two masses are separated 20 m on a horizontal track. At time t = 0, one mass A m

Question

Two masses are separated 20 m on a horizontal track. At time t = 0, one mass A moves towards mass B with an initial velocity of 20 m/s. At the same time, mass B moves with an initial velocity of 15 m/s in the same direction (away from mass A). The coefficient of kinetic friction between each mass and the horizontal surface is 0.36.

a) when do the masses collide?

b) What are the speeds of the two masses just before they collide?

c) If the blocks stick together after the collision, how long is it before they come to stop?

Two masses are separated 20 m on a horizontal track. At time t = 0, one mass A moves towards mass B with an initial velocity of 20 m/s. At the same time, mass B moves with an initial velocity of 15 m/s in the same direction (away from mass A). The coefficient of kinetic friction between each mass and the horizontal surface is 0.36. when do the masses collide? What are the speeds of the two masses just before they collide? If the blocks stick together after the collision, how long is it before they come to stop?

Explanation / Answer

Nice problem.

each block will decelerate due to frictional force. Frictional force depends on mass through the normal force. Deceleration also varies with mass. These two instances of mass cancel out, which is why we can solve (first parts) without knowing the actual masses.

F = ma

FN mu = ma

(9.8 m/s^2)(Mass)(0.36) = (Mass)a

-3.528 m/s^2 = a (call plus the direction both blocks are moving)

a) the nice thing about masses cancelling is that they are slowing down at the same rate, so , as long as they are both moving, their relative speed is always 5 m/s.

So every second, Mass A gets 5m closer to Mass B

Distance = rate*time = d = rt

t = d/r = 20 m / 5 m/s = 4 seconds (still need to check if they are both moving at t = 4)

Check is 15m/s > 4s*3.528m/s^2 ???

yes 14.112m/s is how much speed they have both lost in 4s

Answer 4 seconds until they collide

b) 20 m/s - 14.112m/s = 5.9 m/s for mass A

15m/s - 14.112m/s = 0.9 m/s for mass B

c) if they stick together it is a totally inelastic collision. Ma(5.888) + Mb(0.888) = (Ma + Mb) Vf

the speed depends on the relative masses of A and B, which I call  Ma and Mb

general answer

Vf/3.528m/s^2 = t

where Vf = {Ma(5.888m/s) + Mb(0.888)}/(Ma + Mb)

Let's say they are equal masses, then the velocity just after collision is 6.776 M/2M = 3.388m/s

and the speed slows down 3.528 m/s each second

3.888m/s / 3.528m/s^2 = 0.96 s

So, if equal masses, just about a second later they will come to a stop.

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