Question

Ball a of mass 1kg moving with velocity of 2m /s strikes directly on a ball of mass 2kg rest. What are the velocities of the two balls after impact if coefficient of restitution is 0.5

Answer 1

Explanation:

A ball of mass m1 = 1 kg moving with a velocity u1 = 2 m /s, strikes directly on another ball of mass m2 = 2 kg at rest (u2 = 0). If the co-efficient of restitution between the two balls e = 0.5 . calculate the velocities v1 and v2 of the two balls after impact

Solution. = m1 u1 + m2u2 = m1 u1 + m2u2

1(2) + 2 (0) = 1 (v1) +2 (v2)

or v1 + 2v2 =2

given e =0.5 = v2 – v1 / u1 –u2 = u1 –u2 /2-0

v2 – v1

By adding equations (1) and (2), we get

3v2 = 3

or v2 = 1m/s.Ans

equation (1)

v1 = 2- 2v2 = 2-2 x 1 = 0 m/s.Ans

Answer 2

Answer:

$v_{A}=0 \mathrm{~m} / \mathrm{s}$ and $v_{B}=1 \mathrm{~m} / \mathrm{s}$  are the velocities of the two balls after impact if the coefficient of restitution is 0.5.

Explanation:

For a collision between two objects, the coefficient of restitution exists as the ratio of the relative speed after to the relative speed before the collision. The coefficient of restitution exists as a number between 0 (perfectly inelastic collision) and 1 (elastic collision) inclusive.

Coefficient of restitution = Relative velocity after collision / Relative velocity before collosion

$=\frac{v_{B}-v_{A}}{u_{A}-u_{B}} \ldots . .(1)$

moment conservation:

$\mathrm{m}_{\mathrm{A}} \mathrm{u}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}} \mathrm{u}_{\mathrm{B}}=\mathrm{m}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}} \ldots \ldots \ldots \ldots \text { (2) }$

Given:

$\mathrm{m}_{\mathrm{A}}=1 \mathrm{~kg},$

$\mathrm{u}_{\mathrm{A}}=2 \mathrm{~m} / \mathrm{s},$

$\mathrm{m}_{\mathrm{B}}=2 \mathrm{~kg},$

$\mathrm{u}_{\mathrm{B}}=0 \mathrm{~m} / \mathrm{s},$

e = 0.5

To find:

The velocities of the two balls after impact if the coefficient of restitution is 0.5.

Now, $0.5=\frac{v_{B}-v_{A}}{2-0}$

$\Rightarrow \mathrm{V}_{B}-\mathrm{V}_{\mathrm{A}}=1 \ldots \ldots \ldots(3)$

From moment conservation,

$&1 \times 2+2 \times 0=1 \times v_{A}+2 \times v_{B}$

$&v_{A}+2 v_{B}=2 \ldots \ldots \ldots \ldots . \ldots(4)\end{aligned}$

From equations 3 and 4,

$v_{A}=0 \mathrm{~m} / \mathrm{s}$ and $v_{B}=1 \mathrm{~m} / \mathrm{s}$.

$v_{A}=0 \mathrm{~m} / \mathrm{s}$ and $v_{B}=1 \mathrm{~m} / \mathrm{s}$  are the velocities of the two balls after impact if the coefficient of restitution is 0.5.

Therefore, the correct answer is $v_{A}=0 \mathrm{~m} / \mathrm{s}$ and $v_{B}=1 \mathrm{~m} / \mathrm{s}$.

SPJ3