1. Two railraod cars start from rest rolling down an inclined track. At the bott

Question

1. 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 0.71 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.)

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 0.121 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.)

3. A 4-g bullet is fired into a 1.5-kg block that is resting on a table. After impact, the block with the embedded bullet slides 4 cm across the table before coming to rest again. The coefficient of kinetic friction between the block and the surface of the table is 0.62. 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.)

4. A 3-g bullet is fired into a 1.3-kg pendulum of length 79 cm. The pendulum swings by 14 degrees. What was the initial speed of the bullet?

5. Claudia and Marco are standing face-to-face on two skateboards. They both have equal mass of 78 kg (including the skateboards). Claudia throws a 8.3-kg bowling ball to Marco.
The ball is moving at a speed of 4.1 m/s between the students (ignore that the ball will slighly drop. Also ignore any friction on the skateboards.)

What is Claudia's speed after having thrown the ball?

What is Marco's speed after having caught the ball?

You should have noticed that Claudia's and Marco's speeds are similar but not equal. Why is that?

Below answer with Marco's speed

Explanation / Answer

1) energy of the two cars on the ramp = 2mgh

energy after three cars start moving together at 0.91 m/s will be = 0.5*3m*0.91*0.91 = 1.24215m

on equating the two energy we get 2m*9.8*h = 1.24215m

or h = 63.375 cm

2) energy when the three cars move up the incline is = 3mg*0.121 = 3.5574m

so, 0.5mv2 = 3.5574 m

or v= sqrt(7.1148) = 2.67 m/s

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