In this section, all collisions will be completely elastic – that is, the total

Question

In this section, all collisions will be completely elastic – that is, the total kinetic energy of the system will be constant throughout. Also, the collisions are entirely one-dimensional, with all motion taking place in either the northward or southward direction. 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 mass 3.7 kg, head-on, and rebounds back to the south. What will be the speed of the 1-kg block after this collision? What will be the speed of the second (heavier) block after the collision? Now the first (1-kg) block, after this first collision, 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? What will be the speed of the second block after this second collision? 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 second 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 collisions in total (including the first two described above) will there have been between the two blocks?

Explanation / Answer

Let v1 be the velocity of the 1kg block and v2 be the velocity of the 3.7kg block after collision

Conserving momentum before and after collision,

(1)(5) = (1)(-v1) + (3.7)(v2) [ considered velocity towards north to be positive and south to be negative]

=> v1 = 3.7v2 - 5 ...............(1)

Conserving kinetic energy before and after collision,

(1/2)(1)(52) = (1/2)(1)(v12) + (1/2)(3.7)(v22)

=> 25 = (v12) + 3.7(v22) ..............(2)

Substitutig (1) in (2)

25 = (3.7v2 - 5)2 + 3.7(v22)

=> 25 = (13.69v22 - 37v2 + 25) + 3.7v22

=> v2 = 2.13 m/s

v1 = 3.7v2 - 5

=> v1 = 2.88 m/s

After the 1kg block hits the southern wall and rebound, its velocity will be 2.88 m/s in the north direction

Let v'1 be the velocity of the 1kg block southwards and v'2 be the velocity of the 3.7kg block northwards after collision for the second time

Conserving momentum before and after collision,

(1)(2.88) + (3.7)(2.13) = (1)(-v'1) + (3.7)(v'2) [ considered velocity towards north to be positive and south to be negative]

=> v'1 = 3.7v'2 - 10.76 ...............(1)

Conserving kinetic energy before and after collision,

(1/2)(1)(2.882) + (1/2)(3.7)(2.132) = (1/2)(1)(v'12) + (1/2)(3.7)(v'22)

=> 25.08 = (v'12) + 3.7(v'22) ..............(2)

Substitutig (1) in (2)

25.08 = (3.7v'2 - 10.76)2 + 3.7(v'22)

=> 25.08 = (13.69v'22 - 79.62v'2 + 115.78) + 3.7v'22

=> 17.39v'22 - 79.62v'2 + 90.7 = 0

=> v'2 = 2.45 m/s

v1 = 3.7v2 - 10.76

=> v'1 = -1.7 m/s

The negative sign indicates that the assumed direction for v'1 is wrong . Therefore, the 1kg block will move northwards after collision.

Since, both the blocks are moving northwards after collision, and v'2 > v'1, the 1kg block will not collide with the 3.7kg block any further.

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