A particle of mass 'm' is moving with speed '2v' and collides with a mass '2m' moving with speed 'v' in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass 'm', which move at angle 45° with respect to the origianl direction. The speed of each of the moving particle will be :-

A ball is thrown vertically up (taken as +z-axis) from the ground. The correct momentum-height (p-h) diagram is :

A body of mass 2 kg makes an eleastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body ?

A body of mass m₁ moving with an unknown velocity of $${v_1}\mathop i\limits^ \wedge $$, undergoes a collinear collision with a body of mass m₂ moving with a velocity $${v_2}\mathop i\limits^ \wedge $$ . After collision, m₁ and m₂ move with velocities of $${v_3}\mathop i\limits^ \wedge $$ and $${v_4}\mathop i\limits^ \wedge $$ , respectively. If m₂ = 0.5 m₁ and v₃ = 0.5 v₁, then v₁ is :-