A 25kg crate is at rest on a floor but is pushed by a force of 60N to the right . If the crate remains at rest, determine the co
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Answer:
The ball experiences the greater momentum change
Explanation:
The momentum change of each object is given by:
where
m is the mass of the object
v is the final velocity
u is the initial velocity
Both objects have same mass m and same initial velocity u. So we have:
- For the ball, the final velocity is
Since it bounces back (so, opposite direction --> negative sign) with same speed (so, the magnitude of the final velocity is still u). So the change in momentum is
- For the clay, the final velocity is
since it sticks to the wall. So, the change in momentum is
So we see that the greater momentum change (in magnitude) is experienced by the ball.
Answer:
The ball will drop 0.881 m by the time it reaches the catcher.
Explanation:
The position of the ball at time "t" is described by the position vector "r":
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
Where:
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
When the ball reaches the catcher, the position vector will be "r final" (see attached figure).
The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.
Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):
x = x0 + v0x · t
17.0 m = 0 m + 40.1 m/s · t
t = 17.0 m/ 40. 1 m/s = 0.424 s
With this time, we can calculate the y-component of the vector "r final", the drop of the ball:
y = y0 + v0y · t + 1/2 · g · t²
Initially, there is no vertical velocity, then, v0y = 0.
y = 1/2 · g · t²
y = -1/2 · 9.8 m/s² · (0.424 s)²
y = -0.881 m
The ball will drop 0.881 m by the time it reaches the catcher.
