Please help. No idea how to answers these physics questions teacher never went over it.?

Question

Problems: http://img410.imageshack.us/img410/6060/scalepi6.gif
These are probably trivial but I have no idea how they are derived to the point of solution

#1 is not 4
#2 is right
#3 is not 2

Answer 1

Number 1....

What will the scale read? Anytime you step on a scale, it reads off the normal force for your situtation. So, as the ball sits comfortable on top of the spring, what is the normal force it experiences?

To do this, we need to know what forces are acting on the ball. Thankfully, everything in the y-direction, so here we go.

Forces acting on the ball/cup system...
Weight = -(mb + mc + mp) g
Spring = kx
Normal = W

The weight is negative since it acts straight down.
The x in the spring force is the distance the spring is stretched, which in the problem is given as d.

Now sum up for the forces and set them equal to zero (the ball's not accelerating).

-(mb + mc + mp) g + kx + W = 0
W + kx = (mb + mc + mp) g
W = (mb + mc + mp) g - kx

x = d = (mb g / k)

W = (mb + mc + mp) g - k (mb g / k)
W = (mb + mc + mp) g - mb g
W = mb g + mc g +mp g - mb g
W = mc g + mp g
W = (mc + mp) g

Answer: Choice 4


Problem number 2...
The fact the ball collides inelastically with the cup tells you that the ball and cup suck together throughout the compression. It also tells you that KE wasn't conserved in the collision, but that won't apply to this question.

Again, we need to know the forces.

Weight = -(mb + mc + mp) g
Spring = kx
Normal = W
Impact Force

Unfortunately, we don't know the time it takes for the system to come to rest, so I'm not sure how to calculate the impact force. I may think of it and come back.

Answer 2

They are not trivial; they require some thought. For the first part, #4 applies -- before the ball is added to the cup: the scale weighs the platform and the cup. For the second part, I like #2. For the third part, I also like #2.