ooO T-Mobile LTE4:40 PM Homework Two railraod cars start from rest rolling down
ID: 1467102 • Letter: O
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ooO T-Mobile LTE4:40 PM Homework 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 three cars have the same mass. Ignore any friction STION 2 A railraod 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 O.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 QUESTION 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.) QUESTION 4 A5-g bullet is fired into a 1.3-kg pendulum of length 75 , The pendulum swings by 13 degrees. 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.)Explanation / Answer
3)(m*v bullet) = (m*v block + bullet).
4g = 0.004kg.
11cm = 0.11m.
Mass of block + bullet = (1.1 + 0.004) = 1.104kg.
(1.104 x 9.8) = normal force of 10.8192N.
Friction force = (10.8192 x 0.73) = 7.898N.
Acceleration = (f/m) = 7.898/1.104, = -7.154m/sec^2.
Initial velocity of block = s(2ad) = 1.2545m/sec.
Now apply the algebraic expression above.
(1.2545 x 1.104)/0.004 = bullet velocity of 346.242m/sec.
4) Vertical height gain of pendulum = 0.75m-(cos 13 x 0.75) = 0.0192 metre.
GPE at 13 deg. deflection = (mgh) = (1.3 + 0.005kg) x 9.8 x 0.0192 = 0.24555 Joule. (The bullet mass 0f 0.005kg. is now added to the pendulum mass).
This PE is KE at the bottom of the pendulum swing.
V after bullet lodges in pendulum = (2KE/m)=0.613451m/sec.
So here we go. Initial bullet V = (1.305kg x 0.613451)/0.005 = 160.11m/sec.
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