The 45 kg child in the sled shown starts with initial velocity 2.1 m/s and reach
ID: 1603731 • Letter: T
Question
The 45 kg child in the sled shown starts with initial velocity 2.1 m/s and reaches the bottom of the 9.4 m high hill with a velocity of 8.6 m/s. Find the amount of energy lost due to frictional effects during the slide. A 26 ton locomotive coasting at 3.3 m/s couples with a 6.4 ton passenger car initially at rest. Find the final speed of the train. A bowling ball rolling down the gutter at 2.6 m/s suffers a perfectly elastic collision with a second ball initially at rest Assuming that both balls weigh the same find the final speed of the second ball.Explanation / Answer
7) energy lsot = changein mechanical energy
= (mgh +0.5mv^2)top - (0.5 mv^2) bottom
m = 45 kg, vt = 2.1 m/a, vb = 8.6 m/s
= (45*9.8*9.4 +0.5*45*2.1^2) - (0.5*45*8.6^2)
energy lost = 2580 J
8) m1 = 26 ton , m2 = 6.4 ton , u1 = 3.3 m/s ,u2 =0
from conservation of momentum
m1u1 +m2u2 = (m1+m2)v
26*3.3 +0 = (26+6.4)*v
v = 2.65 m/s
9) m1 = m2 = m , u1 = 2.6 m/s , u2 = 0
from conservation fo momenum
m1u1 +m2u2 = m1v1 +m2v2
2.6 + 0 = v1 +v2 ..(1)
for elastic collision
v2 - v1 = u1 - u2
v2 - v1 = 2.6 ... (2)
from (1) and (2) we get
v1 =0
final speed of second ball v2 = 2.6 m/s
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