question 2-B center of mass problem and mu m = squareroot 3/12, respectively. (g

Question

question 2-B

center of mass problem

and mu m = squareroot 3/12, respectively. (g = 10 m/s^2) (a) When t = 0.8 s, how far is the brick from its original position? (b) When t = 0.8 s, what is the kinetic energy of the brick? (c) What is the loss of mechanical energy between t = 0 s and t = 0.8 s? Suppose that a particle of mass m_1 = squareroot 2 g, moving with initial speed v_1 = 8 times 10^5 m/s, undergoes an elastic collision with a second particle of mass m_2, which is initially at rest. (a) [m_2 = m_1]: In the rest frame, a particle is discovered at an angle 60 degree to the initial direction of m_1. Find the direction and speed of the other particle after the collision. (b) Calculate the directions and speeds of two particles in the above problem (a) in the center of mass frame. (c) [m_2 = 1 g]: In the rest frame, m_1 moves off at an angle theta to its initial direction of motion after the collision. Find the maximum value of sin^2 theta. A particle of mass m is constrained to move on the frictionless inner surface of a cone of half-angle theta.

Explanation / Answer

The speeds will be same as v1 and v2.

Given that m1=1.414g

v1=8*10^5m/s

m2=m2

v2=0

according to law of conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

m1(u1-v1)=m2(v2-u2)

given that m1=m2

1.414(u1-8*10^5)=1.414(0-u2)

u1+u2=8*10^5------(1)

for elastic collisions the valid relation u1-u2=v2-v1-------------(2)

from equations1 and 2

u1=8*10^5m/s

u2=0m/s

and the directions is also the same.

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