4). The setting: a room with a frictionless floor. A block of mass 1 kg is place

Question

4). The setting: a room with a frictionless floor. A block of mass 1 kg is placed on the floor and initially pushed northward, whereupon it begins sliding with a constant speed of 5 m/s. It eventually collides with a second stationary block of a mass of 6.1 kg head on and rebounds back south.

a). What will be the speed of the 1 kg block after this collision?

b) What will be the speed of the second (heavier) block after the collision?

c).Now the first 1kg block, after this first collison, slides south until it collides elastically with the southern wall of the room, bouncing off back to the north. The second block is still sliding northward at this point along the frictionless floor, at the same velocity imparted to it in part (b) above. Eventually the 1 kg block again collides with the second block. What will be the speed of the 1 kg block after this second collision.

d). What will be the speed of the second block after the second collision?

e). Assuming the room in question is very big, there is plenty of room for the blocks to both slide as far north as you might like. As you can imagine, for a while the 1 kg block will keep bouncing off the second block, then back off the southern wall, then back off the seconf block, and so on. But eventually, the 1 kg block will bounce off the southern wall with too low a speed to overtake the second block and there will be no more collisions, At this point how many collsions in total (including the first two described above) will there have been between the two blocks.

Explanation / Answer

initial velocity of 1 kg block, v1i = 5 m/s ( +ve toward north and -ve toward south)

of 6.1kg block, v2i = 0

final velocities, v1f = -v1

v2f = v2

for elastic collision,

velocity of approach = velocity of speration

5 = v1 + v2

v2 = 5 - v1

Applying momentum conservation,

1 x 5 + 0 = -v1 + 6.1v2

putting v2 in terms of v1

5 = - v1 + 6.1*5 - 5v1

v1 = 4.25 m/s ...........Ans(A)

b) v2 = 5 - v1 = 0.75 m/s

c) now 1kg is moving with 4.25 toward north and 6.1kg is moving 0.75 toward north.

4.25 - 0.75 = v2 + v1

v2 = 3.5 - v1

4.25 + 6.1*0.75 = v1 + 6.1v2

8.825 = - 5.1v1 + 3.5*6.1

v1 = 2.45 m/s ....south

d) v2 = 1.05 m/s ....north

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.