A block of mass 1kg is pushed on a movable wedge ofmass 2kg and height h=30 cm with a velocity u=6m//sec. Before striking the wedge it travels 2m on a rough horizontal portion. Velocity is just sufficient for the block to reach the top of the wedge. Assuming all surfaceare smooth except the given horizontal part and collision of block and wedge is jerkless, the friction coefficient of the rough horizontal part is : An observer having a gun observes a remotely controlled balloon. When he first noticed the balloon, it was at an altitude of 800m and moving vertically upward at a constant velocity of 5m//s. The horizontal displacement of balloon from the observer is 1600m. Shells fired from the gun have an initial velocity of 400 m//s at a fixed angle theta ( sin theta =3//5 and cos theta =4//5 ). The observer having gun waits ( for some time after observing the balloon) and fires so as to destroy the balloon. Assume g=10m//s^(2). Neglect air resistance.

A block of mass 1kg is pushed on a movable wedge ofmass 2kg and height

1.	A block of mass 1kg is pushed on a movable wedge ofmass 2kg and height h=30 cm with a velocity u=6m//sec. Before striking the wedge it travels 2m on a rough horizontal portion. Velocity is just sufficient for the block to reach the top of the wedge. Assuming all surfaceare smooth except the given horizontal part and collision of block and wedge is jerkless, the friction coefficient of the rough horizontal part is : An observer having a gun observes a remotely controlled balloon. When he first noticed the balloon, it was at an altitude of 800m and moving vertically upward at a constant velocity of 5m//s. The horizontal displacement of balloon from the observer is 1600m. Shells fired from the gun have an initial velocity of 400 m//s at a fixed angle theta ( sin theta =3//5 and cos theta =4//5 ). The observer having gun waits ( for some time after observing the balloon) and fires so as to destroy the balloon. Assume g=10m//s^(2). Neglect air resistance.
Answer» `0.125` `0.377` `0.675` `0.45` Solution :velocity of the block after passing through the ROUGH SURFACE is `v=sqrt(36+2mug(2))=sqrt(36-40mu)` Apply WORK energy THEOREM `mu mg (2)+mgh=KE_(i)-KE_(f)...(1)` at the highest point `V_(block)=V_(wedg e)` `2-mu+3=(1)/(2)1(6)^(2)-(1)/(2)3v^(2)` `mu=(54)/(80)=0.675`