A block of mass m = 2.08 kg slides down an = 25.2 ° incline which is h = 3.78 m

Question

A block of mass m = 2.08 kg slides down an = 25.2 ° incline which is h = 3.78 m high. At the bottom, it strikes a block of mass M = 7.40 kg which is at rest on a horizontal surface, as seen in figure below. (Assume a smooth transition at the bottom of the incline.) If the collision is elastic, and friction can be ignored,

1) determine the speed of the smaller block after the collision.

2)Determine the speed of the bigger block after the collision.

3)Determine how far back up the incline the smaller mass will go.

Explanation / Answer

for the smaller block , using conservation of energy between Top and Bottom of the incline

Kinetic energy at the bottom = Potential energy at the top

(0.5) m v2 = mgh

v = sqrt(2gh)

v = sqrt(2 x 9.8 x 3.78) = 8.61 m/s

for the collision between the two blocks

vi = initial velocity of smaller block before collision = v = 8.61 m/s

Vi = initial velocity of larger block before collision = 0 m/s

vf = final velocity of smaller block after collision = ?

Vf = final velocity of larger block after collision

using conservation of momentum

m vi + M Vi = m vf + M Vf

(2.08) (8.61) + (7.40) (0) = (2.08) vf + (7.40) Vf

Vf = 17.91 - (2.08) vf                                                  eq-1

using conservation of kinetic energy

m v2i + M V2i = m v2f + M V2f

(2.08) (8.61)2 + (7.40) (0)2 = (2.08) v2f + (7.40) V2f

using eq-1

(2.08) (8.61)2 + (7.40) (0)2 = (2.08) v2f + (7.40) (17.91 - (2.08) vf )2

vf = 7.56 m/s

2)

using eq-1

Vf = 17.91 - (2.08) (7.56)

Vf = 2.18 m/s

3)

the smaller block continue to move in the same direction as before collision and does not travel back up on incline

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