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Conservation of linear momentum and impulse

Two particles of masses M and 2 M, moving as shown, with speed

Question

Hard

Solving time: 4 mins

Two particles of masses $M$ and $2 M$ , moving as shown, with speeds of $10 m / s$ and $5 m / s$ , collide elastically at the origin. After the collision, they move along the indicated directions with speed $v_{1}$ and $v_{2}$ are nearly

(a) $6.5 m / s$ and $3.2 m / s$
(b) $3.2 m / s$ and $6.3 m / s$
(d) $3.2 m / s$ and $12.6 m / s$
(d) $6.5 m / s$ and $6.3 m / s$

Views: 5,419 students

Updated on: Apr 2, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

Found 2 tutors discussing this question

Daniel Discussed

Two particles of masses

M

and

2 M

, moving as shown, with speeds of

10 m / s

and

5 m / s

, collide elastically at the origin. After the collision, they move along the indicated directions with speed

v_{1}

and

v_{2}

are nearly

(a)

6.5 m / s

and

3.2 m / s

(b)

3.2 m / s

and

6.3 m / s

(d)

3.2 m / s

and

12.6 m / s

(d)

6.5 m / s

and

6.3 m / s

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

Exp. (d)
The given condition can be drawn as shown below
Solution Image

Applying linear momentum conservation law in

x

-direction, we get
Initial momentum = Final momentum

(M \times 10 cos 3 0^{\circ}) + (2 M \times 5 cos 4 5^{\circ})

= (M \times v_{2} cos 4 5^{\circ}) + (2 M \times v_{1} cos 3 0^{\circ})

\Rightarrow (M \times 10 \times \frac{3}{2}) + (2 M \times 5 \times \frac{1}{2})

= (M \times v_{2} \times \frac{1}{2}) + (2 M \times v_{1} \times \frac{3}{2})

\Rightarrow 53 + 52 = \frac{v _{2}}{2} + v_{1} 3 \dots (i)

Similarly, applying linear momentum conservation law in

y

-direction, we get

(M \times 10 sin 3 0^{\circ}) - (2 M \times 5 sin 4 5^{\circ})

= (M \times v_{2} sin 4 5^{\circ}) - (2 M \times v_{1} sin 3 0^{\circ})

\Rightarrow (M \times 10 \times \frac{1}{2}) - (2 M \times 5 \times \frac{1}{2})

= (M \times v_{2} \times \frac{1}{2}) - (2 M \times v_{1} \times \frac{1}{2})

\Rightarrow 5 - 52 = \frac{v _{2}}{2} - v_{1} \dots (ii)

Subtracting Eq. (ii) from Eq. (i), we get

(53 + 52) - (5 - 52)

= (\frac{v _{2}}{2} + v_{1} 3) - (\frac{v _{2}}{2} - v_{1})

\Rightarrow 53 + 102 - 5 = v_{1} 3 + v_{1}

\Rightarrow v_{1} = (\frac{5 3 + 10 2 - 5}{1 + 3})

= \frac{8.66 + 14.142 - 5}{1 + 1.732}

= \frac{17.802}{2.732} \Rightarrow v_{1} = 6.516 m / s = 6.5 m / s \dots (iii)

Substituting the value from Eq. (iii) in Eq. (i), we get

53 + 52 = \frac{v _{2}}{2} + 6.51 \times 3

\Rightarrow v_{2} = (53 + 52 - 6.51 \times 3) 2

\Rightarrow v_{2} = (8.66 + 7.071 - 11.215) 1.414

\Rightarrow v_{2} = 4.456 \times 1.414 \Rightarrow v_{2} = 6.3 m / s

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Question Text	Two particles of masses $M$ and $2 M$ , moving as shown, with speeds of $10 m / s$ and $5 m / s$ , collide elastically at the origin. After the collision, they move along the indicated directions with speed $v_{1}$ and $v_{2}$ are nearly (a) $6.5 m / s$ and $3.2 m / s$ (b) $3.2 m / s$ and $6.3 m / s$ (d) $3.2 m / s$ and $12.6 m / s$ (d) $6.5 m / s$ and $6.3 m / s$
Updated On	Apr 2, 2023
Topic	Laws of motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 3
Upvotes	401
Avg. Video Duration	13 min

Text solutionVerified

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