Question

A Ikg ball moving at 12 ms-' collides headon with a 2 kg ball moving in the opposite direction at
PLATINUM PACKAGE
[C]
2
24 ms. If the coefficient of restitution is
3
b) 120 J
the energy lost in the collision is
d) 48 J
a) 601
c) 240 J
67​

Answer 1

★ Question :

A 1 kg ball is moving at 12 m/s collides head on with a 2kg ball moving in the opposite direction with at 14 m/s . If the coefficient of restitution is 2/3 , then how energy is lost in collision ?

★ Solution :

Given ,

Mass of 1st ball , m₁ = 1 kg

Initial velocity of 1st ball , u₁ = 12 m/s

Final velocity of 1st ball , v₁ = v₁ m/s

Mass of 2nd ball , m₂ = 2 kg

Initial velocity of 2nd ball , u₂ = - 14 m/s

Final velocity of 2nd ball , v₂ = v₂ m/s

Apply Conservation of momentum ,

➠ Initial momentum = Final momentum

➠ $\rm m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

➠ $\rm (1)(12) + (2) (-14) = (1)v_1 + (2) v_2$

➠ $\rm 12 - 28 = v_1 + 2 v_2$

➠ $\rm -16 = v_1 + 2v_2$

➠ $\rm v_1 + 2v_2 = - 16\ \; \dashrightarrow\ (1)$

$\rm Coefficient\ of\ restitution= \dfrac{2}{3}$

➠ $\rm \dfrac{v_2 - v_1}{u_1 - u_2} = \dfrac{2}{3}$

➠ $\rm 3 v_2 - 3 v_1 = 2 u_1 - 2 u_2$

➠ $\rm 3v_2 - 3 v_1 = 24 + 28$

➠ $\rm 3v_2 - 3v_1 = 52\ \; \dashrightarrow\ \; (2)$

On Solving both equations , we get ,

$\rm \pink{v_1 = - \dfrac{152}{9}\ \; ,\ \; v_2 = \dfrac{4}{9}}\ \; \bigstar$

$\longrightarrow \rm v_1 =- 16.89\ m/s\ ,\ v_2 = 0.45\ m/s$

Energy lost in collision :

$\longrightarrow \rm \Delta K_1 - \Delta K_2$

$\longrightarrow \rm \dfrac{1}{2} \big\{ m_1 u_1 ^2 +m_2 u_2 ^2 -m_1 v_1 ^2 - m_2 v_2 ^2 \big\}$

$\longrightarrow \dfrac{1}{2} \big\{ (1)(12)^2 + (2)(14)^2 -(1)(-16.89)^2-(2)(0.45)^2 \big\}$

$\longrightarrow \rm \dfrac{1}{2} \big\{ 144 +392-285.27 - 0.405 \big\}$

$\longrightarrow \textsf{\textbf{125.16\ J}}\ \; \orange{\bigstar}$

Energy lost in the collision is 125 J (approx.) .

Answer 2

Answer:

Given that :-

Mass of 1st ball , m₁ = 1 kg

Initial velocity of 1st ball , u₁ = 12 m/s

Final velocity of 1st ball , v₁ = v₁ m/s

Mass of 2nd ball , m₂ = 2 kg

Initial velocity of 2nd ball , u₂ = - 14 m/s

Final velocity of 2nd ball , v₂ = v₂ m/s

Need to Find :-

Energy lost in collision

Solution :-

$\sf \: m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

(1)(12) + (2)(-14) = (1)v₁ + (2)v₂

12 + (-28) = v₁ + 2v₂

12 - 28 = v₁ + 2v₂

-16 = v₁ + 2v₂ (EQ 1)

Now,

Coefficient = ⅔

v₂ - v₁/u₂ - u₁ = ⅔

3v₂ - 3v₁ = 2u₂ - 2u₁

3v₂ - 3v₁ = 52

3v₁ - 3v₂ = 52

Now,

v₁ = 6.23 m/s

v₂ = -11.12 m/s

Energy Lost = ∆K₁ - ∆K₂

= ½[(1)(12)² + (2)(14)² - (1)(-16.89)² - 2(0.45)²]

= ½ [144 + (2)(14)² - (1)(-16.89)² - 2(0.45)²]

= ½ [144 + 392 - (1)(-16.89)² - 2(0.45)²]

= ½ [144 + 392 - 285.27 - 0.405]

= ½ [536 - 285 - 0.405]

= ½ × 250.595

≈ 125 J