A flat, thin sheet of metal (m 1 =8.63 kg) is initially at rest on a slippery, l

Question

A flat, thin sheet of metal (m1=8.63 kg) is initially at rest on a slippery, level surface that is essentially frictionless: it's smooth, wet ice. The level upper surface of the metal sheet is smooth but not frictionless. At time t=0, a small bag of sand (m2=1.24 kg), which was previously a projectile, lands on the sheet, without bouncing or breaking, and then slides across the sheet (without bouncing of breaking) until it comes to rest on the sheet. SHown here is the graph of vi*x(t), the x-velocity of the metal sheet for the first 15 seconds after impact.

The impact speed of the sandbad on the metal was 19.0 m/s. Ignore air drage, and assume g=9.80 m/s/s.

b)Calculate the coefficient of static friction between the metal sheet and the sandbag.

c)What was the peak height of the sandbag - above the level of the metal sheet - before its impact?

A flat, thin sheet of metal (m, = 863 kg) is initially at reston slippery, level surface that is essentially frictionless it's smooth, wet ice. The level upper surface of the metal sheet is smooth but not frictionless. e (mis 1.70 4, Al tine 0. a small bag of sand (m, = 1.24 kg), which was previoasly a projectile, lands on the sheet, without bouncing or breaking, and then slides across the sheet (without bouncing or breaking) until it comes to rest on the sheet. Shown here is a graph of v,(0) the r-velocity of the metal sheet for the first 15 seconds after the impact. 10 15 The impact speed of the sandbag on the metal was 19.0 m/s. Ignore air drag, and assume g 980 ml a. Draw a time graph of v,, (the r-velocity of the sandbag) for the time interval 0 sIs 15s b. Calculate the coefficient friction ent of kinetic friction, , between the metal sheet and the sandhag What was the peak height of the sandbag-above the level of the metal sheet-before its impact? c.

Explanation / Answer

As there is no frcition between metal plate and surface below it, momentum of bag + plate in horizontal direction is conserved. let speed of bag in horzontal direction, before hitting the plate be V2xo. Since final terminal velocity of plate is 1.7 m/s, that is the final common velocity of plate and bag. By monetum conservation in horizontal direction,
1.24*V2xo = (1.24 + 8.63)*1.7
V2xo = 13.53 m/sec.

a) As there is constant force of friction acting on the bag, it deccelrates uniformly. Hence graph of v2x against time will be straight line upto t = 10 sec, joining points
(0,13.53) and (10,1.7)
After 10 sec, it will horizontal line.

b) Accelration of plate = 1.7 - 0/10 = 0.17
force on plate = mu m2 g
so it's accelration = mu m2 g / m1
0.17 = mu 1.24*9.8/8.63
mu = 0.12

c) Velocity of bag in vertical direction, just before hitting the plate V2yo = root(V22-V2xo2)
Maximum height of bag = V2yo2 / 2g = (V22-V2xo2)/2g = 9.1 m

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