Question

A 6.0 kg block is sliding on a leve, frictionless surface at a speed of 5.0 m/s when it undergoes a head-on, perfectly inelastic collision with a 4.0 kg block that is initially at rest on the top of a frictionless, 2.0 m high inclined plane. A) What is the speed of the combined blocks when they reach the bottom of the incline? B) If the ground at the bottom of the incline is level, and if the coefficient of kinetic friction between the combined blocks and the ground is 0.80, how much further beyond the incline will the tandem of blocks slide before they come to rest?

Answer #1

according to conservation of momentum

m1u1 + m2u2 = (m1 + m2) v

or, 6.0x5.0 + 4.0x0 = (6.0 +4.0) v

or, v = 3.0 m/s

so the after collision the combined block velocity = 3.0 m/s

now the potential energy of the combined block at a height of 2.0 m = mgh = 10 x9.8x2.0 = 196 J

now this amount of energy will convert into kinetic energy when the combine d block strike on ground

let v1 be the speed of the combined blocks when they reach the bottom of the incline

so 0.5Mv1^2 = 196

or, 0.5 x(m1 + m2) x v1^2 = 196

or. 0.5x10 x v1^2 = 196

or, v1 = 6.26 m/s ..........................................................ans

b)

let the distance be s upto which the blocks slide before they come to rest

so, The frictional force is mg * 0.80

so the acceleration is mg* 0.80 / m = g * 0.80

and from v^2 = 2as

or, 6.26^2 = 2 x 9.8x0.80 x s

or, s = 2.499 m ..............................................ans

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