
A ball is allowed to fall freely from a height of 3 meters onto a fixed plate. The successive rebound heights are ${h_1},{h_2},{h_3},.......$. If the distance covered by the ball before coming to rest is y meters, find y. (Given that coefficient of restitution is 0.5)
Answer
407.7k+ views
Hint: This question is based on the concepts of motion. We need to understand the equations of motion and their applications for a freely falling body. The coefficient of restitution is the ratio of final velocity to initial velocity for two objects after they collide. The restitution coefficient, denoted by the letter ‘e,' is a unitless quantity with values ranging from 0 to 1.
Complete step by step answer:
Let the velocity with which the ball strikes the plate be ${v_1}$. Let ${v_2}$ be the velocity of rebound. Let the height from which the ball is dropped be $H$.
From the equations of motion, we know that,
${v_1} = \sqrt {2gH} $
Here $g$ is acceleration due to gravity.
${v_2} = \sqrt {2g{h_1}} $
$
e = \dfrac{{{v_2}}}{{{v_1}}} = \sqrt {\dfrac{{{h_1}}}{H}} \\
{e^2} = \dfrac{{{h_1}}}{H} \\
{h_1} = {e^2}H \\
$
Similarly, we get,
${h_2} = {e^2}{h_1} = {e^4}H$and so on.
Therefore, total distance before it comes to stop,
$
y = H + 2{h_1} + 2{h_2} + 2{h_3} + ..... \\
= H + 2{e^2}H + 2{e^4}H + 2{e^6}H + ..... \\
= H[1 + 2{e^2}(1 + {e^2} + {e^4} + .....)] \\
= H[1 + \dfrac{{2{e^2}.1}}{{1 - {e^2}}}] \\
= H[\dfrac{{1 + {e^2}}}{{1 - {e^2}}}] \\
= 3[\dfrac{{1 + {{0.5}^2}}}{{1 - {{0.5}^2}}}] \\
= 5m \\
$
Thus, the ball will cover a distance of 5 meters before coming to rest.
Note: The coefficient of restitution gives us knowledge about the collision's elasticity. A perfectly elastic collision is described as one in which no overall kinetic energy is lost. The maximum coefficient of restitution for this form of collision is e = 1. A perfectly inelastic collision is one in which all of the kinetic energy is lost.
Complete step by step answer:
Let the velocity with which the ball strikes the plate be ${v_1}$. Let ${v_2}$ be the velocity of rebound. Let the height from which the ball is dropped be $H$.
From the equations of motion, we know that,
${v_1} = \sqrt {2gH} $
Here $g$ is acceleration due to gravity.
${v_2} = \sqrt {2g{h_1}} $
$
e = \dfrac{{{v_2}}}{{{v_1}}} = \sqrt {\dfrac{{{h_1}}}{H}} \\
{e^2} = \dfrac{{{h_1}}}{H} \\
{h_1} = {e^2}H \\
$
Similarly, we get,
${h_2} = {e^2}{h_1} = {e^4}H$and so on.
Therefore, total distance before it comes to stop,
$
y = H + 2{h_1} + 2{h_2} + 2{h_3} + ..... \\
= H + 2{e^2}H + 2{e^4}H + 2{e^6}H + ..... \\
= H[1 + 2{e^2}(1 + {e^2} + {e^4} + .....)] \\
= H[1 + \dfrac{{2{e^2}.1}}{{1 - {e^2}}}] \\
= H[\dfrac{{1 + {e^2}}}{{1 - {e^2}}}] \\
= 3[\dfrac{{1 + {{0.5}^2}}}{{1 - {{0.5}^2}}}] \\
= 5m \\
$
Thus, the ball will cover a distance of 5 meters before coming to rest.
Note: The coefficient of restitution gives us knowledge about the collision's elasticity. A perfectly elastic collision is described as one in which no overall kinetic energy is lost. The maximum coefficient of restitution for this form of collision is e = 1. A perfectly inelastic collision is one in which all of the kinetic energy is lost.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Trending doubts
10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Define least count of vernier callipers How do you class 11 physics CBSE

The combining capacity of an element is known as i class 11 chemistry CBSE
