When riding your dirt bike you run into a tree which causes your bike to immediately come to a stop. You fly over the handle bar
2 answers:
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A) Car A is initially ahead
B) The two cars are at the same point at the times: t = 0, t = 2.27 s and
t = 5.73 s
C) The distance between the two cars is not changing at t = 1.00 s and t = 4.33 s
D) The two cars have same acceleration at t = 2.67 s
Explanation:
A)
The position of the two cars at time t is given by the following functions:
with
Substituting,
And
with
Substituting,
Here we want to find which car is ahead just after they leave the starting point. To find that, we just need to calculate the position of the two cars after a very short amount of time, let's say at t = 0.1 s. Substituting this value into the two equations, we get:
So, car A is initially ahead.
B)
The two cars are at the same point when their position is the same. Therefore, when
which means when
Re-arranging the equation, we find
One solution of this equation is t = 0 (initial point), while we have two more solutions given by the equation
which has two solutions:
t = 2.27 s
t = 5.73 s
So, these are the times at which the cars are at the same point.
C)
The distance between the two cars A and B is not changing when the velocities of the two cars is the same.
The velocity of car A is given by the derivative of the position of car A:
The velocity of car B is given by the derivative of the position of car B:
Therefore, the distance between the two cars is not changing when the two velocities are equal:
This is another second-order equation, which has two solutions:
t = 1.00 s
t = 4.33 s
D)
The acceleration of each car is given by the derivative of the velocity of the car A.
The acceleration of car A is:
While the acceleration of car B is:
So, the two cars have same acceleration when
And solving the equation, we find:
So, the two cars have same acceleration at t = 2.67 s.
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(a) 1.43 m/s
We can solve this problem by using the law of conservation of energy.
The initial total energy stored in the spring-mass system is
where
k = 7.91 N/m is the spring constant
Substituting,
The final kinetic energy of the ball is equal to the energy released by the spring + the work done by friction:
where
is the kinetic energy of the ball, with
being the mass of the ball
v being the final speed
is the work done by friction (which is negative since the force of friction is opposite to the motion), where
is force of friction
d = 14.5 cm = 0.145 m is the displacement
Substituting,
So, the kinetic energy of the ball as it leaves the cannon is
And so the final speed is
(b) +5.08 cm
The speed of the ball is maximum at the instant when all the elastic potential energy stored in the spring has been released: in fact, after that moment, the spring does no longer release any more energy, so the kinetic energy of the ball from that moment will start to decrease, due to the effect of the work done by friction.
The elastic potential energy of the spring is
And this has all been released when it becomes zero, so when x = 0 (equilibrium position of the spring). However, the spring was initially compressed by 5.08 cm, so the ball has maximum speed when
x = +5.08 cm
with respect to the initial point.
(c) 1.78 m/s
The maximum speed is the speed of the ball at the moment when the kinetic energy is maximum, i.e. when all the elastic potential energy has been released.
As we calculated in part (a), the total energy released by the spring is
E = 0.0102 J
The work done by friction here is just the work done to cover the distance of
d = 5.08 cm = 0.0508 m
Therefore
So, the kinetic energy of the ball at the point of maximum speed is
And so the final speed is