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 4.2 kg, head-on, and rebounds back to the south. (a) What will be the speed of the 1-kg block after this collision? Answer: __________.___ m/s (b) What will be the speed of the second (heavier) block after the collision? Answer: __________.___ m/s (c) 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? Answer: ________.___ m/s (d) What will be the speed of the second block after this second collision? Answer: ___________.___ m/s (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 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? Answer: ___________ collisions

Explanation / Answer

initial speed v = 5 m/s north

after first collision, Speed of the big block be V1 towards north and that of small block be v1 towards south,

Applying conservation of momentum,

1x5 = 4.2 V1 - 1 x v1 => 5 = 4.2V1 - v1 eqn 1

but since it is elastic collision, velocity of approach = velocity of separation,

So, 5 = V1 + v1 eqn 2

Solving eqns 1 and 2,

we get V1 = 1.923 m/s north (ans to part b )

and v1 = 3.077 m/s (ans to part a)

After bouncing back from the wall, the small block will have the same speed v1 in the north direction.

again applying conservation of momentum,

1x 3.077 + 4.2 x 1.923 = 4.2 V2 - v2

11.183 = 4.2 V2 - v2 eqn 3

Also, again it is an elastic collision. So, velocity of approach = velocity of separation

v1 - V1 = V2 + v2 = >   3.077 - 1.923 = V2 + v2

1.154 = V2 + v2 eqn 4

Solving eqns 3 and 4 , we get

V2 = 2.373 m/s ( ans to part d )

v2 = - 1.219 m/s (ans to part c)

- ve sign means that, unlike expected, the smaller mass is also moving towards north after the second collision with speed 1.219 m/s which is less than the speed of bigger block (2.373) . So, they will never collide again.

e.) So, total number of collisions = 2

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