1. The Hulk, with a mass of 200 kg, falls off a 5 m tall building and lands on t

Question

1. The Hulk, with a mass of 200 kg, falls off a 5 m tall building and lands on the raised side of a seesaw. At the other end of the seesaw that is initially resting on the ground is a rock with a mass of 5.0 kg. Once the Hulk’s side hits the ground and the seesaw stops moving, it launches the rock towards the Hulk’s direction at an angle of 45° with respect to the ground. When the rock leaves the seesaw, it does so 1 m above the ground. If the building the Hulk fell off of is 3.2 m away, how high up does the rock smash into the building?

2.A 1 kg block is released from rest at the top of a frictionless ramp that is in the shape of a quarter of a circle and is 6/g m high (here I’m using g only for its value and not its units). At the end of the ramp, the block simultaneously hits a spring with a 2 N/m spring constant, and a surface with a 1/g coefficient of kinetic friction (again I’m using g for its value only). If the block comes to rest after compressing the spring by an unknown amount, what is the total distance traveled by the block?

I am sorry that there is no picture for these questions.

Since I need to know how to solve them, please show me all of steps.

Explanation / Answer

1.) We need to note the following points so as to solve the given problem:

a.) As the see-saw will launch the rock, the velocity with which it will start to move will be same as the velocity with which Hulk will hit the ground, as they are on the same see-saw. The speed with which one end will go down will be same as the speed with which the other end will go up.

b.) The rock while moving towards the building will suffer acceleration only in the vertical direction and will be equal to g.

c.) We can conserve, Hulk's energy to determine the speed which which he hits the ground and sets the rock in motion

We will use the above to solve the given problem.

For Hulk's motion, we have:

MgH = 0.5 M v^2

or V = sqrt (2gH) = 9.905 m/s

Now we also know that the rock leaves the see-saw at the speed of 9.9.05 m/s at an angle of 45 degrees with the ground. Also the building is 3.2 m away.

Time taken by the rock to cover the horizontal distance to the building would be:

t = 3.2 / Vcos45 [Distance along horizontal divided by horizontal component of velocity]

t = 0.4569 seconds

Now, the height at which the rock will be at this point of time, will be the height at which it will hit the building.

S = Ut - 0.5gt^2

S = 9.905 Sin45 (0.4569) - 0.5 x 9.81 x (0.4569)2

S = 3.201 - 1.024 = 2.177 metres.

Therefore the required height at which the rock hits the building = 3.177 metres [Since it was launched from 1 m above the ground]

2.) As the block slides down the ramp and then hits the spring placed on surface with friction, we will use conservation of energy to determine the distance it travels on the flat surface agains the spring.

For the block sliding down the ramp from 6/g metres we can write:

mgh = 0.5 m v^2

or 6 = 0.5 v^2

or v = sqrt12 = 3.464 m/s

Now this kinetic energy will get dissipated by the friction and the spring's potential energy.

Hence we can again write

6 = 0.5 * 2 * x^2 + 1g * (1/g) x [Since kinetic energy = Friction force x distance + Spring's PE]

or, 6 = x^2 + x

or x = 2 or -3 [Solve the quadratic equation to get the values of x as 2 or -3 now,- 3 cannot be a value hence 2 m is the required values]

Therefore the block travels for 2 metres against the spring.

Total distance travelled = Distance along the ramp + 2metres

Total distance = 3.14 x 6/2g + 2 = 2.96 metres

NOTE: You need to note that the quadrant would be such placed that the two ends of the curved section will make an angle of 90 degrees with the horizontal, hence the height would be equal to radius.

Also, you can directly equate the PE at the top with the work done by friction and PE of spring. You don't need to determine the KE at the bottom of the ramp as I have done. I did that only to simplify things. Plus, such approach can potentially help with other problems.

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