A particle of mass m is projected from the ground with an initial speed `u_0` at an angle `alpha` with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed `u_0.` The angle that the composite system makes with the horizontal immediately after the collision is

Options

A. ` pi //4`
B. ` (pi//4) + ajpga`
C. ` (pi //2)- alpha`
D. (pi//2)`

Answer

A

Explanation

Refer to Fig. 2 (CF). 64 `,
. ltBrgt speed of the first particle at highest jpoint (A),
` u_1 = u_0 cos alph`
speed of the second particle at the highest point
` A` si ` u_2 = sqrt ( u_0^2 -2 gH)`
where ` H= (i_0^2 sin ^2 alpha)/( 2g)` ltbRgt :. ` u_2 = sqrt (u_0^2 - 2 gu_0^2 sib=n^2 alpha//2 g) = u_0 cos alpha`.
Let ` hat i` and ` hat j` be the unit vectors along horizontal and vertical direction respectively . Then
` ve u_1 = u_0 cos alpha hat i ` and vec u_2 = u_0 cos alpha that j`
Let vec v) be the velocity of composite system after inelastic collision and ` theta ` is the andle which it makes with the horixontal direction . Accordign to law if consrevation of linear momentum
` 2 m ( v_x hat i + v_y hat j) = mu_0 coa alpha hat i+ mu_0 cos alph hat j`
We have , ` v_x = ( u_0 cos alpha )/2 and ` v_y = ( u_0 cos alph)/2`
Then have , ` v_y/-x = ((u_0 cos alpha)//2)/( ( u_0 cos alpha)//2) = 1 or ` theta = (pi)/4`.