
A metal ball falls from a height of \[32{\text{ }}m\] on a steel plate. If the coefficient of restitution is \[0.5\;\] to what height will the ball rise after a second bounce?
A. ${\text{2 }}m$
B. ${\text{4 }}m$
C. ${\text{8 }}m$
D. ${\text{16 }}m$
Answer
164.7k+ views
Hint: In this question, we have a metal ball that is dropped from a height of \[32{\text{ }}m\]and the ball reaches on the steel plate and rebounds. We need to find the rise (height) of the steel ball when it rebounds. We’ll use the formula for velocity of a body falling from a height $H$ and a relation between coefficient of restitution and velocity. Substituting all the given values in these formulas we will get the required solution.
Formula used:
If the ball is falling from the height $H$, then its velocity is;
$v = \sqrt {2gH} $
Relation between velocity and coefficient of restitution –
${v_f} = ev$
Complete step by step solution:
Given that,
Steel ball dropped from a height \[32{\text{ }}m\].
Let, the height be $H = 32{\text{ }}m$
Now, the ball reaches the plate and bounces from the steel plate. Let the height to which the steel ball rebounds be $h$. As we know, If the ball is falling from the height $H$, then its velocity is
$v = \sqrt {2gH} $
Put $H = 32{\text{ }}m$
Velocity of steel ball when it strikes the plate will be $v = \sqrt {64g} $
After striking the plate, the steel ball rebounds with the velocity ${v_f}$.
Where, ${v_f} = ev$ ----------(1)
Here, $e = $ coefficient of restitution and $v = $ velocity of striking the plate
Also, final velocity will be ${v_f} = \sqrt {2gh} $ ---------(2)
From equation (1) and (2), we get
$\sqrt {2gh} = ev$
It can be written as $\sqrt {2gh} = e\sqrt {64g} $
Squaring both the sides of the above equation
It will be $2gh = 64g{e^2}$
$ \Rightarrow h = 32{e^2}$
Put \[e = 0.5\;\] (given)
It implies that, $h = 8{\text{ }}m$
Hence, option C is the correct answer i.e., ${\text{8 }}m$.
Note: When two bodies collide, the coefficient of restitution is the ratio of the two objects' final relative velocity to their starting relative velocity following contact. The coefficient of restitution of a bouncing ball is found by taking the square root of the ratio of one bounce to the height of the following bounce.
Formula used:
If the ball is falling from the height $H$, then its velocity is;
$v = \sqrt {2gH} $
Relation between velocity and coefficient of restitution –
${v_f} = ev$
Complete step by step solution:
Given that,
Steel ball dropped from a height \[32{\text{ }}m\].
Let, the height be $H = 32{\text{ }}m$
Now, the ball reaches the plate and bounces from the steel plate. Let the height to which the steel ball rebounds be $h$. As we know, If the ball is falling from the height $H$, then its velocity is
$v = \sqrt {2gH} $
Put $H = 32{\text{ }}m$
Velocity of steel ball when it strikes the plate will be $v = \sqrt {64g} $
After striking the plate, the steel ball rebounds with the velocity ${v_f}$.
Where, ${v_f} = ev$ ----------(1)
Here, $e = $ coefficient of restitution and $v = $ velocity of striking the plate
Also, final velocity will be ${v_f} = \sqrt {2gh} $ ---------(2)
From equation (1) and (2), we get
$\sqrt {2gh} = ev$
It can be written as $\sqrt {2gh} = e\sqrt {64g} $
Squaring both the sides of the above equation
It will be $2gh = 64g{e^2}$
$ \Rightarrow h = 32{e^2}$
Put \[e = 0.5\;\] (given)
It implies that, $h = 8{\text{ }}m$
Hence, option C is the correct answer i.e., ${\text{8 }}m$.
Note: When two bodies collide, the coefficient of restitution is the ratio of the two objects' final relative velocity to their starting relative velocity following contact. The coefficient of restitution of a bouncing ball is found by taking the square root of the ratio of one bounce to the height of the following bounce.
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