Question

An explosion in a rigid pipe shoots three balls out of its ends. A 6 gam ball comes out the right end. A 4 gram ball comes out the left end with twice the speed of the 6 gram ball. From which end does the third ball emerge?

Answer 1

Answer:

The third ball emerges from the right side.

Explanation:

This is a conservation of Momentum problem

In an explosion or collision, the momentum is always conserved.

Momentum before explosion = Momentum after explosion

Since the rigid pipe was initially at rest,

Momentum before explosion = 0 kgm/s

- Taking the right end as the positive direction for the velocity of the balls

- And calling the speed of the 6 g ball after explosion v

- This means the velocity of the 4 g ball has to be -2v

- Mass of the third ball = m

- Let the velocuty of the third ball be V

Momentum after collision = (6)(v) + (4)(-2v) + (m)(V)

Momentum before explosion = Momentum after explosion

0 = (6)(v) + (4)(-2v) + (m)(V)

6v - 8v + mV = 0

mV - 2v = 0

mV = 2v

V = (2/m) v

Note that since we have established that the sign on m and v at both positive, the sign of the velocity of the third ball is also positive.

Hence, the velocity of the third ball according to our convention is to the right.

Hope this Helps!!!

Answer 2

Answer:

the third ball emerge from right end .

Explanation:

initially the three balls are at rest so u = 0  

Given m1 = 6g

v1 = v(i) (Right end indicates positive X axis so it is represented as (i))

m2 = 4g

v2 = 2(v)(-i) (left end indicates negative X axis so it is represented as (-i))

m3 = m

v3 = v'

Applying conservation of momentum

m1(u1) + m2(u2) + m3(u3) = m1(v1) + m2(v2) + m3(v3)

u1 = u2 = u3 = u = 0 ( at rest)

0 = (6×10⁻³)(v)(i) + (4×10⁻³)(2v)(-i) + m(v')

m(v') = (2×10⁻³)(v)(i)

v' = (2×10⁻³)(v)(i)/(m)

So positive (i) indicates right end

So the third ball emerge from right end .