Ball A of mass m_1 moving with a velocity v under goes a head-on collision with

Question

Ball A of mass m_1 moving with a velocity v under goes a head-on collision with ball B of mass m_2 at rest. After collision, ball A continues moving in its original direction with half its original sped. The speed of ball B after collision will be m_1 v/2m_2 m_2 v/2m_1 m_1 v/m_2 m_2 v/m_1 A shell of mass 2m fired with a speed u at an angle theta to the horizontal explodes at the highest point of its trajectory into two fragments of mass m each. If one fragment falls vertically, the distance at which the other fragment falls from the gun is given by u^2 sin 2 theta/g 3 u^2 sin 2 theta/2g 2 u^2 sin 2 theta/g 3 u^2 sin 2 theta/g A truck carrying sand is moving on a smooth horizontal road with a uniform sped u. If a mass Delta m of sand leaks in a time Delta t from the bottom of the truck, the force needed to keep the

Explanation / Answer

Here ,

as the centre of mass will fall at the same place if there were no explosion

position of first stone on falling , x1 = u^2 * sin(2 * theta)/2g

Now , position of centre of mass , Xcom = u^2 * sin(2*theta)/g

for the position of farther particle x2

(x1 + x2)/2 = Xcom

u^2 * sin(2 * theta)/2g + x2 = 2 * u^2 * sin(2*theta)/g

x2 = 3*u^2 * sin(2*theta)/2g

the correct option is b) 3*u^2 * sin(2*theta)/2g

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