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Chapter 4: Problem 521

Four identical balls are lined in a straight grove made on a horizontal frictionless surface as shown. Two similar balls each moving with a velocity v collide elastically with the row of 4 balls from left. What will happen (A) One ball from the right rolls out with a speed \(2 \mathrm{v}\) and the remaining balls will remain at rest. (B) Two balls from the right roll out speed \(\mathrm{v}\) each and the remaining balls will remain stationary. (C) All the four balls in the row will roll out with speed \(\mathrm{v}(\mathrm{v} / 4)\) each and the two colliding balls will come to rest. (D) The colliding balls will come to rest and no ball rolls out from right.

Short Answer

Expert verified
The correct answer is Option B: Two balls from the right roll out with a speed \(v\) each, and the remaining balls remain stationary. This scenario satisfies both the conservation of momentum and the conditions for an elastic collision.

Step by step solution

01

Define the known variables

We are given: 1. Identical balls 2. Horizontal frictionless surface 3. Two balls moving with initial velocity v collide elastically with a row of four balls.
02

Conservation of Momentum

The conservation of momentum states that the total initial momentum equals the total final momentum. Let's calculate the total initial momentum. Total Initial Momentum = (Mass of colliding balls) x (Initial Velocity) Total Initial Momentum = 2m(v) Now, let's look at each option.
03

Option A: One ball from the right rolls out with a speed 2v

If this option were true, the momentum after the collision would be: Total Final Momentum = (Mass of rolling ball from the right) x (Final Velocity) = m(2v) However, this does not satisfy the conservation of momentum because 2m(v) ≠ m(2v). So, option A is incorrect.
04

Option B: Two balls from the right roll out with speed v each

If this option were true, the momentum after the collision would be: Total Final Momentum = (Mass of rolling balls from the right) x (Final Velocity) = 2m(v) This option does satisfy the conservation of momentum since 2m(v) = 2m(v). Let's now check if the conditions of an elastic collision are met.
05

Checking the conditions for an elastic collision

Now let's analyze option B in terms of an elastic collision. In an elastic collision, the relative speed of approach is equal to the relative speed of separation. In our case the relative speed of approach is 2v - 0 = 2v (subtracting the speeds of the moving balls and the stationary row of 4 balls). In option B, two balls from the right roll out with speed v each and the remaining balls are stationary. So the relative speed of separation is also 2v since both balls are moving with velocity v. Since the relative speed of approach is equal to the relative speed of separation, option B fulfills the conditions for an elastic collision.
06

Conclusion

Option B is the correct answer. Two balls from the right roll out with a speed v each, and the remaining balls remain stationary. This scenario satisfies both the conservation of momentum and the conditions for an elastic collision.

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Most popular questions from this chapter

A spring with spring constant \(\mathrm{K}\) when stretched through $2 \mathrm{~cm}\( the potential energy is \)\mathrm{U}\(. If it is stretched by \)6 \mathrm{~cm}$. The potential energy will be...... (A) \(6 \mathrm{U}\) (B) \(3 \mathrm{U}\) (C) \(9 \mathrm{U}\) (D) \(18 \mathrm{U}\)

A bullet of mass \(\mathrm{m}\) moving with velocity \(\mathrm{v}\) strikes a block of mass \(\mathrm{M}\) at rest and gets embedded into it. The kinetic energy of the composite block will be (A) \((1 / 2) \mathrm{mv}^{2} \times[\mathrm{M} /(\mathrm{m}+\mathrm{m})]\) (B) \((1 / 2) \mathrm{mv}^{2} \times[(\mathrm{m}+\mathrm{m}) / \mathrm{M}]\) (C) \((1 / 2) \mathrm{MV}^{2} \times[\mathrm{m} /(\mathrm{m}+\mathrm{M})]\) (D) \((1 / 2) \mathrm{mv}^{2} \times[\mathrm{m} /(\mathrm{m}+\mathrm{M})]\)

From an automatic gun a man fires 240 bullet per minute with a speed of $360 \mathrm{~km} / \mathrm{h}\(. If each weighs \)20 \mathrm{~g}$, the power of the gun is (A) \(400 \mathrm{~W}\) (B) \(300 \mathrm{~W}\) (C) \(150 \mathrm{~W}\) (D) \(600 \mathrm{~W}\)

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