Answer:For a perfectly elastic collision, the final velocities of the carts will each be 1/2 the velocity of the initial velocity of the moving cart. For a perfectly inelastic collision, the final velocity of the cart system will be 1/2 the initial velocity of the moving cart.
Explanation:
Answer:
a) 39.6 m/s b) 4123 N
Explanation:
a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).
Fnet=ma
ma=m(v^2/R) (centripetal acceleration)
mg=m(v^2/R)
m cancels out (this is why pilot feels weightless) so,
g=(v^2/R)
9.8 m/s^2 = v^2/160 m
v^2=1568 m^2/s^2
v=39.6 m/s
b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.
Convert 300 km/hr to m/s
300 km/hr=83.3 m/s
Convert pilot's weight into mass:
760 N = 77.55 kg
Fnet=ma
n-mg=m(v^2/R)
n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)
n=3363.2 N+760 N=4123 N
Answer:
the ball's velocity was approximately 0.66 m/s
Explanation:
Recall that we can study the motion of the baseball rolling off the table in vertical component and horizontal component separately.
Since the velocity at which the ball was rolling is entirely in the horizontal direction, it doesn't affect the vertical motion that can therefore be studied as a free fall, where only the constant acceleration of gravity is affecting the vertical movement.
Then, considering that the ball, as it falls covers a vertical distance of 0.7 meters to the ground, we can set the equation of motion for this, and estimate the time the ball was in the air:
0.7 = (1/2) g t^2
solve for t:
t^2 = 1.4 / g
t = 0.3779 sec
which we can round to about 0.38 seconds
No we use this time in the horizontal motion, which is only determined by the ball's initial velocity (vi) as it takes off:
horizontal distance covered = vi * t
0.25 = vi * (0.38)
solve for vi:
vi = 0.25/0.38 m/s
vi = 0.65798 m/s
Then the ball's velocity was approximately 0.66 m/s
The speed is changing its direction all the time. There
is an acceleration which changes the direction of the speed – that is called
centripetal acceleration. Only uniform linear motions are considered to have no
acceleration.
This is the general formula for acceleration
a = dv/dt
When calculating dv, you should keep in mind the change
in the velocity vector’s direction. You can easily see in a graph that with dt
tending to 0 (so the length of the arc covered is also tending to 0), the difference
between vectors Vf and V0 has a direction which is perpendicular to velocity
(the shorter the arc, the closest the angle is to 90 degrees).
There is a formula (which can be deducted from the
previous formula) which allows you to calculate the acceleration:
a = v^2/r
Let’s talk about the units:
v is in m/s
r is in m
so v^2/r
is in (m/s)^2/m = (m^2/s^2)/m = m/s^2
which is the same unit as dv/dt:
dv/dt = (m/s)/s= m/s^2
First, foremost, and most critically, you must look at the graph, and critically
examine its behavior from just before until just after the 5-seconds point.
Without that ability ... since the graph is nowhere to be found ... I am hardly
in a position to assist you in the process.
