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Standard XII

Physics

Momentum

Three particl...

Question

Three particles of mass $20 g, 30 g$ and $40 g$ are initially moving along the positive direction of the three coordinate axes $x, y$ and $z$ respectively with the same of $20 c m / s$ . If due to their mutual interaction, the first particle comes to rest, the second acquires a velocity $^1 0 i +^2 0 k .$ what is the velocity (in $c m / s$ ) of the third particle?

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Solution

$m_{1} = 20 g$ and $u_{1} = 20 i$
$m_{2} = 30 g$ and $u_{2} = 20 j$
$m_{3} = 40 g$ and $u_{3} = 20 k$

$v_{1} = 0$ and $v_{2} = 10 i + 20 k$

From conservation of momentum since no external force is acting on the system-

$m_{1} u_{1} + m_{2} u_{2} + m_{3} u_{3} = m_{1} v_{1} + m_{2} v_{2} + m_{3} v_{3}$

$⟹ 400 i + 600 j + 800 k = 0 + 300 i + 600 k + 40 v_{3}$

$⟹ 40 v_{3} = 100 i + 600 j + 200 k$

$⟹ v_{3} = (2.5 i + 15 j + 5 k) c m / s$

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m1=20g and u1=20im2=30g and u2=20jm3=40g and u3=20kv1=0 and v2=10i+20k From conservation of momentum since no external force is acting on the system-m1u1+m2u2+m3u3=m1v1+m2v2+m3v3⟹400i+600j+800k=0+300i+600k+40v3⟹40v3=100i+600j+200k⟹v3=(2.5i+15j+5k)cm/s