Basketball and tennis ball-2!

Classical Mechanics Level 3

Now consider $n$ balls, $B_1, . . . , B_n$ , having masses $m_1, m_2, . . . , m_n$ $(with~ m_1 >>m_2 >>· · ·>>m_n)$ , sitting in a vertical stack. The bottom of $B_1$ is a height $h$ above the ground, and the bottom of $B_n$ is a height $h + l$ above the ground. The balls are dropped. In terms of $n$ , to what height does the top ball bounce?

(Note: Work in the approximation where $m_1$ is much larger than $m_2$ , which is much larger than $m_3$ , etc., and assume that the balls bounce elastically.)

Try part 1 also - Basketball and tennis ball!

Question Source - Basketball and tennis ball

H = l + (4^n)^2 h

H = l + (2^n)^2 h

H = l + (4^n -1)^2 h

None of These

H = l + (2^n -1)^2 h

1 solution

Nishant Rai
May 20, 2015

Just before $B_1$ hits the ground, all of the balls are moving downward with speed $v =\sqrt{2gh}$ .

We will inductively determine the speed of each ball after it bounces off the one below it. If $B_i$ achieves a speed of $v_i$ after bouncing off $B_{i-1}$ , then what is the speed of $B_{i+1}$ after it bounces off $B_i$ ?

The relative speed of $B_{i+1}$ and $B_i$ (right before they bounce) is $v + v_i$ .

This is also the relative speed after they bounce. Since $B_i$ is still moving upwards at essentially speed $v_i$ , the final upward speed of $B_{i+1}$ is therefore $(v + v_i) + v_i$ .

Thus, $v_{i +1} = 2v_i + v ..... (3)$ .

Since $v_1 = v$ , we obtain $v_2 = 3v$ , $v_3 = 7v$ , $v_4 = 15v$ , etc.

In general, $v_n = (2n - 1)v ....... (4)$ which is easily seen to satisfy eq. (3), with the initial value $v_1 = v$ .

From conservation of energy, $B_n$ will bounce to a height of $H = l +\frac{((2n - 1)v)^2}{2g} = l + (2n - 1)^2h$

hey friend,

in the alternatives there is l+(2^n-1)²h, and not the correct answer, that is l+(2n-1)²h.

by the way, i get the incorrect answer because of this. I marked "none of these".

Claudio Felipe - 6 years ago

@Claudio Felipe

this solution is correct !

while writing the answer, i made a mistake of writing $(2n)$ instead of $(2^n)$ . I have now edited the solution!

Nishant Rai - 6 years ago

it's ok thanks

Claudio Felipe - 6 years ago

Basketball and tennis ball-2!

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