A block of mass m = 2.70 kg slides down a 30.0 incline which is 3.60 m high. At

Question

A block of mass m = 2.70 kg slides down a 30.0 incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 7.80 kg which is at rest on a horizontal surface (Figure 1) . (Assume a smooth transition at the bottom of the incline.) The collision is elastic, and friction can be ignored.

Part A.) Determine the speed of the block with mass m = 2.70 kg after the collision.

Part B.)Determine the speed of the block with mass M = 7.80 kg after the collision.

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

Explanation / Answer

An elastic collision means total energy BEFORE collison = total energy AFTER collision. KE remains constant as collision occurs on a horizontal surface.

The smaller block of mass(m) = 2.70 kg slides w/o friction down an incline.
The weight (force of gravity) of this block= mg = (9.81)(2.70) = 26.5 N
The vertical drop of the incline= h = 3.60 sin 30 = 1.80 m
KE of block as it leaves the bottom of incline = its PE at top of incline
KE = PE = mgh = (26,5)(1.80) = 47.7 J = TOTAL energy BEFORE collision.

let V11 = velocity of smaller block BEFORE collsion
KE of smaller block BEFORE = 1/2 m(V11)² = 47.7
(V11)² = 47.7/(0.5)(2.70) = 35.33
V11 = 5.94 m/s {assumed POSITIVE direction}
let V21 = 0 = velocity of larger block BEFORE collision
let V12 = smaller block's velocity AFTER collision.
let V22 = larger block's velcoity AFTER collison

by conservation of energy:
(0.5)(2.70)(V21)² + (0.5)(7.80)(V22)² = 47.7
1.35(V21)² + 3.90(V22)² = 47.7

by conservation of momentum:
(2.70)(V11) + 0 = (2.70)(V21) + (7.80)(V22)
(2.70)(5.94) = 16.0 = (2.70)(V21) + (7.80)(V22)
16 - 2.7V21 = 7.8V22
V22 = 16/7.8 - 2.7/7.8V21 = 2.0512 - 0.346V21

substitute for V22 in energy equation (above):
1.35(V21)² + 3.90[2.0512 - 0.346V21]² = 47.7
1.35(V21)² + 3.90[4.207 - 1.419(V21) + 0.1197(V21)²] = 47.7
1.35(V21)² + 16.4073 - 5.534(V21) + 0.467(V21)² = 47.7
1.81683(V21)² - 5.534(V21) - 31.29 = 0
solve quadratic for V21 get two real roots:
V21 = 5.94 <= X
V21 = - 2.89 <= ANS
accept ONLY the negative value for V21 because implied in the question is the fact that the smaller mass travels back up the incline (negative direction).
or
compute the total KE energy with both sets of values for V21 & V22 to find which set gives a total KE = 47.7 J.
V22 = 2.0512 - 0.346V21  

V22 = 3.0511 <= ANS

A) ANS
V21 = - 2.89 m/s
B) V22 = 3.0511 m/s

C) ANS
KE of smaller mass after collision = (0.5)(2.70)(2.89)² = 11.275 J
This energy converts to PE at max height =11.275/(mg) = 11.275/(2.70)(9.81) = 0.425 m..........Ans.

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