A 2 kg mass is moving towards the right at a speed 3 m/s on a flat, horizontal,

Question

A 2 kg mass is moving towards the right at a speed 3 m/s on a flat, horizontal, frictionless table. It collides perfectly ellastically with a 0.1 kg mass which is initially at rest. The surface of the table is 1.7 m above the floor.

1. After the collision what is the speed of the 2 kg mass?
2. After the collision what is the speed of the 0.1 kg mass?
3. At what distance from the base of the table (directly below the edge of the table's surface) does the 2 kg mass hit the floor?
4. At what distance from the base of the table (directly below the edge of the table's surface) does the 0.1 kg mass hit the floor?

Please show work.

Explanation / Answer

let subscript 1 denote 2kg mass and 2 denote 0.1 kg mass

m1=2 kg, u1=3 m/s, m2=0.1 kg, u2=0

m1u1 + m2u2 = m1v1 + m2v2 ==> 2v1 + 0.1v2 = 6 ....(1)

also, since perfectly elastic collision is occuring,

v2-v1 = u1-u2 ==> v2-v1 = 3 ....(2)

multiplying 2 to eq 2 and adding to 1, we get

2.1 v2 = 12 ==> v2 = 40/7 ==> v1 = 19/7

1. v1 = 2.714 m/s

2. v2 = 5.714 m/s

now, a body with horizontal vel v, from height h, falls on a the ground.

h = gt2/2 ==> t = (2h/g)

and distance from the edge of the table, R = vt = v*(2h/g)

so,

3. R1 = v1*(2h/g) = 1.598 m

4. R2 = v2*(2h/g) = 3.366 m

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