Two railraod cars start from rest rolling down an inclined track At the bottom o
ID: 1467190 • Letter: T
Question
Two railraod cars start from rest rolling down an inclined track At the bottom of the track, they collide and couple with a third car The final speed of the three coupled cars should not exceed 0 72 m/s What is the maximum height from which the two cars can be released? (Assume that all throe cars have the same mass Ignore any friction.) A railiraod car is moving on a level (horizontal) track It collides and couples with two other cars Together the three cars should move up an inclined track to a height of 0.124 m and come to rest What must be the initial speed of the moving car? (Assume that all three cars have the same mass Ignore any friction.) A 4 - g bullet is fired into a 1.1-kg block that is resting on a table After impact, the block with the embedded bullet slides 14 cm across the table before coming to rest again. The coefficient of kinetic friction between the block and the surface of the table is 0.73. What was the initial speed of the bullet? (Practise how to derive an algebraic expression for the speed of the bullet before calculating the final result.) A 5-g budget is fired into a 1 3-kg pendulum of length 75 cm The pendulum swings by 13 degrees What was the initial speed of the bullet? (Practice how to derive an algebraic expression for the speed of the bullet before calculating the final result.)Explanation / Answer
from momentum conservation
afte collision
3*m*V = (m*v1+m*v1)
3*0.72 = 2*v2
v1 = 1.08
initila potentia energy Ui = m*g*h
at the bottom kinetic energy = 0.5*m*v1^2
from energy conservation
0.5*m*v1^2 = m*g*H
H = V1^2/2G = (1.08^2)/(2*9.81) = 0.06 M <<--------ANSWER
++++++++++++++++++++++
after colliison
3*m*v^2 = 3*m*g*h
3*v^2 = 3*9.81*0.124
v = 1.1 m/s
from momentum conservation
momentum before coliision = momentum after coliision
m*V1 = 3*m*V
v1 = 3v = 3.3 m/s
++++++++++
after collision
W = dKE
uk*(M+m)*g*x = 0.5*(M+m)*V^2
0.73*9.81*0.14 = 0.5*v^2
v = 1.42
from momentum conservation
momentum before coliision = momentum after coliision
m*U = (M+m)*V
0.004*u = (1.1+0.004)*1.42
u = 392 m/s
++++++++
potential energy of the pendulum PE = (m+M)*g*L*(1-cos13)
kinetic energy = 0.5*(M+m)*V^2
0.5*(M+m)*V^2 = (m+M)*g*L*(1-cos13)
0.5*V^2 = 9.81*0.75*(1-cos13)
v = 0.61 m/s
from momentum conservation
momentum before coliision = momentum after coliision
m*U = (M+m)*V
0.005*u = (1.3+0.005)*0.61
u = 159.21 m/s
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