5. [lpt] A block of mass m = 2.35 kg slides down an 32.8° incline which is h = 3

Question

5. [lpt] A block of mass m = 2.35 kg slides down an 32.8° incline which is h = 3.71 m high. At the bottom, it strikes a block of mass M 6.66 kg, which is at rest on a horizontal surface, and an elastic collision occurs Assume frictionless surfaces, and a smooth transition at the bottom of the incline. The goal of this problem is to calculate how far back up the incline mass m will travel. To do this, the problem must be broken up into three parts. First, calculate the velocity, v1, of the mass m immediately before it strikes M Answer: Submit All Answers 6. [Ipt] The final velocity of m at the bottom of the ramp that you just calculated now becomes the initial velocity in the collision. Calculate the velocity of m immediately after the collision. Answer: Submit All Answers 7. [1pt] The final velocity immediately after the collision now becomes the initial velocity of the small block sliding back up the ramp. Determine how far (i.e., what distance) back up the incline m will go Answer: Submit All Answers 8. [lpt] If the larger block has a velocity v-2.53 m/s before the collision, what is the final velocity of the smaller block after the collision? (Hint: for this entire problem it is best to work the elastic collision in the centre-of- mass reference frame.) Answer: Submit All Answers

Explanation / Answer

5. Applying energy conservation,

m g h + 0 = 0 + m v1^2 /2

v1 = sqrt(2 g h)

v1 = 8.53 m/s

6. for elastic collision,

Vvelocity of separation = velocity of approach

v + V = 8.53

V = 8.53 - v

Applying momentum conservation,

(2.35 x 8.53) = - 2.35 v + 6.66(8.53 - v )

v = 4.08 m/s

7. Applying energy conservation,

h = v^2 / 2 g = 4.08^2 / (2 x 9.81) = 0.85 m

8. v + V = 8.53 + 2.53

V = 11.06 - v

momentum conservation,

(2.35 x 8.53) - (6.66 x 2.53) = - 2.35 v + 6.66(11.06- v )

v = 7.65 m/s

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