Suppose that a particle of mass m_1 = Squareroot 2g, moving with initial speed v

Question

Suppose that a particle of mass m_1 = Squareroot 2g, 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. [m_2 = m_1]: In the rest frame, a particle is discovered at angle 60 degree to the initial direction of m_1. Find the direction and speed of the other particle after the collision. Calculate the directions and speeds of two particles in the above problem (a) in the center of mass frame. [m_2 = 1g]: In the rest frame, m_1 move off at an angle theta to its initial direction of motion after the collision. Find the maximum value of sin^2 theta.

Explanation / Answer

a) consider elastic collision,

m1=m2= sqrt2 g

it is given that after elastic collision one particle is travelling in the direction of m1 at 60 degree angle

now let us take momentum balance in x axis,

inital momentum = final momentum,

m1*u + m2*0 = m1* v1 cos(60) + m2*v2cos(theta)

=>u= v1cos(60) + v2cos(theta) ( m1= m2) ...............(1)

now ,

along the y axis we have ,

m1*0 + m2*0= m1*v1(sin(60) + m2*v2(sin(theta) .....(2)

now from conservation of kinetic energy,

m1u2 = m1V12 + m2V22 ...............(3)

so the equations are,

8x105 m/s = 0.5V1 +V2 cos(theta)

=> V2 cos(theta)= 8x105 m/s - 0.5V1 ..............(1)

0.866v1 + v2(sin(theta) =0

=>v2(sin(theta) = - (0.866v1 ).............(2)

(8x105 )2= V12 + V22 ................(3)

squaring and adding (1) and (2) we have

V22= (8x105 m/s - 0.5V1 )2 +( - (0.866v1 ))2 ...............(4)

substituting this in (3)

we get

(8x105 )2 = 2V12 -(8x105) V1 +(8x105 )2

=> V1= 0 , or V1= 4x105 m/s

now , V2= 6.92 x105m/s

theta= 30.03 degree

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.