Two point particles collide in a glancing collision. Particle 1 is stationary at

Question

Two point particles collide in a glancing collision. Particle 1 is stationary at the origin and particle 2 is moving at 4m/s on the x-axis prior to the collision. As a result of the collision particle 1 is deflected 30 with respect to the x-axis. Particle 2 is similarly deflected -30 with respect to the x-axis. After the collision the particles are moving at the same speed.

(a) Find the final speeds of both particles.

(b) Find the force particle 2 experiences during the collision of it lasts 0.2s.

Explanation / Answer

conserving momentum after and before collision,

m*(4 i ) = m*(v1*(cos30 i + sin30 j)) + m*(v2*(cos30 i - sin30 j))

So, 4i + 0 j = (v1*cos30 + v2*cos30) i + (v1*sin(30) - v2*sin30) j

Comparing the components on both sides:

4 = (v1*cos30 + v2*cos30) -------(1)

and 0 = v1*sin(30) - v2*sin30 ------(2)

So, v1 = 2.31 m/s

v2 = 2.31 m/s

So, final speeds of both particles = 2.31 m/s

b)

force experiesnced by particle 2 : m*sqrt((4-2.31*cos(30 deg))^2 + (2.31*sin(30 deg))^2)/0.2 = m*2.31/0.2

So, force = 5*m*2.31 = 11.55*m <-------- Just plug in the value of m to get the answer

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