A block of mass 1.58 k g which is initially at rest is acted upon by an impulse

Question

A block of mass 1.58kg which is initially at rest is acted upon by an impulse of 8.22Nm due to a constant force. The impulse lasts for 0.187s seconds and results in the mass moving at velocity v. The block then slides up a frictional ramp which has a length 2.00m and a height 34.8cm before colliding with a mass 3.16kg . In this problem all horizontal surfaces are frictionless (only the incline is frictional).

Part A: With what velocity does the mass move after the impulse acts upon it?

Part B: What average force acts upon the mass during the impulse?

Part C: If the block reaches the top of the incline with velocity 3.91m/s , what is the coefficient of kinetic friction of the ramp?

Part D: If the 1.58kg  block collides and sticks to the 3.16kg  block, what final velocity results?

Part E: How much Kinetic Energy was lost in the collision of the two blocks?

A block of mass 1.58kg which is initially at rest is acted upon by an impulse of 8.22Nm due to a constant force. The impulse lasts for 0.187s seconds and results in the mass moving at velocity v. The block then slides up a frictional ramp which has a length 2.00m and a height 34.8cm before colliding with a mass 3.16kg . In this problem all horizontal surfaces are frictionless (only the incline is frictional). With what velocity does the mass move after the impulse acts upon it? What average force acts upon the mass during the impulse? If the block reaches the top of the incline with velocity 3.91m/s , what is the coefficient of kinetic friction of the ramp? If the 1.58kg block collides and sticks to the 3.16kg block, what final velocity results? How much Kinetic Energy was lost in the collision of the two blocks?

Explanation / Answer

Impulse = change in momentum = 8.22kgm/s = Favg*0.187s => Favg = 43.96N
8.22 = 1.58*(vf-vi) = 1.58(vf-0) => vf = v = 8.22/1.58 = 5.2m/s (1
Favg = 43.96N (2
Angle of ramp = arcsin(0.348/2) = 10 degrees
At the top of the incline PE=mgh=1.58*9.8*0.348 = 5.4J
At the top of the incline KE = 1/2 *m*v^2 = 0.5*1.58*3.91^2 = 12.1J
Total energy at top of incline = KE+PE = 12.1+5.4 = 17.5J
Total energy at bottom of incline = 1/2 *m*v^2 = 0.5*1.58*5.2^2 = 21.4J
Loss = 21.4-17.5 = 3.92J = Ff*x = Ff*2 = m*g*cos10*Cf*2 = 1.58*9.8*cos10*Cf*2 => Cf = 0.13 3)
1.58*3.91 = [1.58+3.16]*vx => vx = 1.30m/s 4)
KEi =1/2 * 1.58*3.91^2 = 12.1J
KEf = 1/2 *[1.58+3.16]*1.30^2 = 4.03J
KEi - KEf = KElost = 8.1J

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