A body A- of mass m-0-1 kg has an initial velocity of 3 - ms-1- It collides elastically with another body- B of the same mass- which has an initial velocity of 5 - ms-1- After collision- A moves with a velocity v -4- - - The energy of B after collision is written as x10 J- The value of x is -

Notations in this question: i= initial ; f= final nbsp; ; A= Body A B= Body B Given: |v_iA|=3 ms^ 1; |v_iB|=5 ms^ 1; |v_fA|=4√(2) ms^ 1 In elastic collision ...

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Standard XII

Physics

COM in Oblique Collision

A body A, of ...

Question

A body $A$ , of mass $m = 0.1 kg$ has an initial velocity of $3^i {ms}^{- 1}$ . It collides elastically with another body, $B$ of the same mass, which has an initial velocity of $5^j {ms}^{- 1}$ . After collision, $A$ moves with a velocity $\to v = 4 (^i +^j)$ . The energy of $B$ after collision is written as $\frac{x}{10} J$ . The value of $x$ is _____.

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Solution

Given:
$| v_{i A} | = 3 {ms}^{- 1}; | v_{i B} | = 5 {ms}^{- 1}; ∣ ∣ v_{f A} ∣ ∣ = 4 \sqrt{2} {ms}^{- 1}$

In elastic collision,
$K E_{i} = K E_{f}$

$\Rightarrow \frac{1}{2} \times m \times 9 + \frac{1}{2} \times m \times 25 = \frac{1}{2} m \times 32 + \frac{1}{2} m v_{f B}^{2}$

$34 = 32 + v_{f B}^{2}$

$\Rightarrow v_{f B} = \sqrt{2}$
$K E_{B} = \frac{1}{2} m v_{f B}^{2} = \frac{1}{2} \times 0.1 \times 2 = 0.1 J$

$\Rightarrow K E_{B} = \frac{1}{10} J$
$∴ x = 1$

Hence, $1$ is the correct answer.

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Given: |viA|=3 ms−1; |viB|=5 ms−1; ∣∣vfA∣∣=4√2 ms−1 In elastic collision, KEi=KEf ⇒12×m×9+12×m×25=12m×32+12mv2fB 34=32+v2fB ⇒vfB=√2 KEB=12mv2fB=12×0.1×2=0.1 J ⇒KEB=110 J ∴x=1 Hence, 1 is the correct answer.