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Grade 12Mechanics

4 identical balls of radius R and mass M are lying in a gravity free space. The balls are in contact and their centres are forming vertices of a square of Side 2R in horizontal plane. One identical ball travelling vertically with speed V hits the four balls symmetrically. The collision is perfectly elastic. The centre of the four stationery balls are at (R,R,0),(R,-R,0),(-R,R,0) and (-R,-R,0).
A. find the speed of ball 1 after collision
B. calculate the speed of ball which was initially traveling with speed V just after collision
C. Find the unit vector in the direction of motion of ball 2 just after collision

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Profile image of pradeep
10 Years agoGrade 12
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1 Answer

Profile image of Tanishq Khare
9 Years ago

Consider the system of 4 balls to be acting on the COM.

Momentum conservation equation along z-axis:---

mv = 4m*v1 - m*v2 ---- (i)

As, the travelling ball will return by the same path due to symmmetry.

e = velocity of separation/velocity of approach

1 = (v1+v2)/v

v = v1+v2 ---- (ii)

Solving 1 & 2

we get,

v2 = 3v/5;

Now apply energy conservation equation before and after collision, as it is elastic collision

energy will be conserved.

consider, speed of 4 similar balls = v3

0.5mv^2 = 0.5*4m*(v3)^2 + 0.5 *m*((3/5)v)^2

m*(16/25)*v^2 = 4m*(v3)^2

v3 = 2v/5;