A particle of mass m1 moving at a speed v1 allong the positive x direction- collides with another particle of mass m2 at rest- After the collision- the particles have velocities v1- and v2- in the xy-plane in the directions of -1 and -2 with the x-axis as shown in the figure- There are no external forces acting on the particles- Express your answer in terms of m1- m2- v1- -1 and -2- What is the total momentum of the system before the collision direction and magnitude-

Total momentum of system before collision =m_1v_1↑ +0 ⇒p⃗ =m_1v_1↑ ⇒( p⃗ #160; ) =m_1v_1 Hence, the answer is m_1v_1. ...

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Standard XII

Physics

Oblique Collision

A particle of...

Question

A particle of mass $m_{1}$ moving at a speed $v_{1}$ allong the positive $x d i r e c t i o n$ , collides with another particle of mass $m_{2}$ at rest. After the collision, the particles have velocities $v_{1}^{'}$ and $v_{2}^{'}$ in the $x y$ -plane in the directions of $θ_{1}$ and $θ_{2}$ with the x-axis as shown in the figure. There are no external forces acting on the particles. Express your answer in terms of $m_{1}, m_{2}, v_{1}, θ_{1}$ and $θ_{2}$ . What is the total momentum of the system before the collision (direction and magnitude)?

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Solution

Total momentum of system before collision
$= m_{1} v_{1} ↑ + 0$
$\Rightarrow \to p = m_{1} v_{1} ↑$
$\Rightarrow (\to p) = m_{1} v_{1}$
Hence, the answer is $m_{1} v_{1} .$

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Total momentum of system before collision=m1v1↑+0⇒→p=m1v1↑⇒(→p)=m1v1Hence, the answer is m1v1.

Total momentum of system before collision
$= m_{1} v_{1} ↑ + 0$
$\Rightarrow \to p = m_{1} v_{1} ↑$
$\Rightarrow (\to p) = m_{1} v_{1}$
Hence, the answer is $m_{1} v_{1} .$