- #1
octu
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Homework Statement
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A 1kg cart, a 0.1kg ball, and an elastic cord connecting the ball and the cart, are placed in a weightless environment; see Figure. The cart has the shape of a symmetric trapezoid with its left and right edges at a 60-deg angle from the vertical. It is also attached with four rollers (two on top and two on bottom) so that it can move freely horizontally. At t = 0, the system is in static equilibrium with the elastic cord fully stretched to exert a force Fs = -1 N (to the left) on the cart. The ball is then released from rest in its t = 0 position and hits the cart at t = 2 seconds. Note that the attachment of the elastic cord is at exactly the same level as the centre of mass of the cart. Answer the following questions. (a) Will the cart move right after the release of the ball (t = 0+)? Provide your reason(s). (b) What are the impulsive forces at the instant the ball hits the cart? (c) Assume the ball sticks to the cart surface, what is the speed of the cart right after the ball hits the cart? Neglect the mass of the string.
M = 1kg
m = 0.1kg
θ = 60
Homework Equations
F = ma
p = mv
The Attempt at a Solution
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a) The cart will want to move after the ball is released because the ball and cart are connected by the cord. The force pulling the ball to the cart is equal and opposite to the force pulling the cart to the ball. However, there is a ledge blocking the cart's motion, so it will stay put.
b) Right of the bat I'm not really sure what's meant by impulsive forces. I thought to find an impulsive force a time frame over which the force acts must be known? Anyways, I figure the ball is pulled in by a force of 1N over two seconds, so the acceleration and velocity after 2s can be given by:
am = 1/m
vm = 2/m
The impulse is given by the change in momentum. The ball goes from vm = 2/m stationary at the moment of contact, so:
Impulse = 2/m
and impulsive force would be this divided by some time? Not sure.
Well, whatever the impulsive force is, it acts normal to the cart surface and can be broken down into x and y components. the x component is proportional to cos(60) and the y component to sin(60).
Another thing - I feel that the vm value is incorrect, because wouldn't the force change over time, if the cord is elastic?
c) The ball has momentum mvm before it hits the cart. A proportion cos(60) of this momentum is carried forward, so:
mvm = (m + M)*v*cos(60)
mvm / (( m + M)*cos(60)) = v
where v is the velocity of the ball and cart stuck together.
Any ideas?