Elastic collision of two particles. The first particle has a of 0.1 kg and has a

Question

Elastic collision of two particles. The first particle has a of 0.1 kg and has a velocity of 1-m/s. The second particle has the same and has the same velocity in the opposite direction. a) What is the velocity of the two particles after the collision (assume that the incoming particle moves to the right)? b) Compute the velocity of the center of mass, before and after the collision. c) Assume that the duration of the collision is 0.01 seconds. What will be the impulse on each particle? d) What will the kinetic energy of each particle before and after the collision? Consider the following torsional pendulum. A wire of length 9.5 m is attached to a mass of m = 1 kg. The mass and the wire form simple pendulum. The pendulum oscillates in the horizontal plane (the plane xy) with a period of T = 2 pi seconds. a) Can you determine the value of the gravitational acceleration. b) Identify the formula torsion-constant kappa? in of the length of the pendulum, the mass and the gravitational acceleration. What will be the value of kappa? c) Assume that for small angle theta (the deviation from equilibrium) the energy conservation is given by: E = E_ + E_p E_kappa = 1/2 omega^2; E = kappa/2 theta^2 I is the moment of inertia, I = mL^2, theta is the angle deviation from equilibrium, omega is the

Explanation / Answer

(2)

Apply conservation of momentum

m1 u1 + m2 u2 = m1 v1 + m2 v2

final velocity of the first ball is

v1 = (m1- m2/ m1+ m2) u1 + 2 m2/ m1 + m2 u2

= 0 + 2 ( 0.1)/(0.1+0.1) * ( -1 m/s)

=-1 m/s

v2 = 2 m1/ m1 + m2 * u1

= 2(0.1)/0.1+0.1 * 1

= 1 m/s

vcm ( before ) = m1 u1 + m2 u2/m1 + m2 = 0.1(1) + 0.1(-1)/0.1+0.1 = 0 m/s

vm ( after) = m1 v1 + m2 v2/m1 + m2 = 0.1(-1) + 0.1(1)/0.1+0.1 = 0 m/s

impuse = change in momentum = m1 ( v1-u1) = 0.1(-1-1) = -0.2 kg m/s

m2 ( v2-u2) = 0.1(1-(-1) = 0.2 kg m/s

kineitc energy before

1/2 m1 u1^2 = 1/2 *(0.1) ( 1^2 = 0.05 J

1/2 m2 u2^2 = 1/2 ( 0.1) (-1)^2 = 0.05 J

after the collision

1/2 m1 v1^2 = 1/2 ( 0.1) ( -1)^2 = 0.05 J

1/2 m1 v2^2 = 1/2 ( 0.1) (1)^2 = 0.05 J

As per guide lines I worked one problem , please post remaining questions in the next post

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