A ball of mass 0.2 ~kg is thrown from a height of 1 ~m and with an initial velocity of 10 ~m / s at an angle of 45^ with the horizontal. Assuming, acceleration due to gravity g=10 ~m / s ^2 , then modulus of momentum increment during the total time of motion in kg m / s is

Step-by-step solution

There is no change in velocity in x -direction, so momentum change occurs only in y -direction. Now initial momentum in y direction is p _y( initial )=m v_y=0.2 ( 10 45^ j ) = l 5 j kg ⁻ ms ⁻¹ Final momentum in y -direction is aligned p y_ final & =m v_y=m ( .u_y^2+2 g h ) . & = (0.2 ( 5 ^2 )+2(-10)(-1) ) j & =-1 j kg - ms ⁻¹ (downwards) aligned So, change in momentum aligned & p=p y_ ( fimal -p x_ initial =-1 j - 1 5 j & =- ( l + l 5 ) j kg ⁻ ms ⁻¹ & =| p|=1+ 1 5 = 1+ 5 5 & aligned