complete question:
A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk
Answer:
F = 1776 N
Explanation:
mass of ball = 60 g = 0.06 kg
velocity of downward direction = 22 m/s = v1
velocity of upward direction = 15 m/s = v2
Δt = 1/800 = 0.00125 s
Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.
p = mv
When a particle move freely and interact with another system within a period of time and again move freely like in this scenario it has a definite change in momentum. This change is defined as Impulse .
I = pf − pi = ∆p
F = ∆p/∆t = I/∆t
let the upward velocity be the positive
Δp = mv2 - m(-v1)
Δp = mv2 - m(-v1)
Δp = m (v2 + v1)
Δp = 0.06( 15 + 22)
Δp = 0.06(37)
Δp = 2.22 kg m/s
∆t = 0.00125
F = ∆p/∆t
F = 2.22/0.00125
F = 1776 N
Answer:
b. 
Explanation:
As we know that the electric field due to infinite line charge is given as
here we can find potential difference between two points using the relation
now we have
now we have
now plug in all values in it
now we know by energy conservation
the answer is (a) molecules
what is the final speed of the incoming ball if it is much more massive than the stationary ball? express your answer using two significant figures. v1 = 200 m / s submitprevious answers correct
Perfectly elastic collisions means that both mechanical energy and
momentum are conserved.
Therefore, for this case, we have the equation to find the final velocity of the incoming ball is given by
v1f = ((m1-m2) / (m1 + m2)) v1i
where,
v1i: initial speed of ball 1.
v1f: final speed of ball 1.
m1: mass of the ball 1
m2: mass of the ball 2
Since the mass of the ball 1 is much larger than the mass of the ball 2 m1 >> m2, then rewriting the equation:
v1f = ((m1) / (m1) v1i
v1f = v1i
v1f = 200 m / s
answer
200 m / s
part b part complete what is the final direction of the incoming ball with respect to the initial direction if it is much more massive than the stationary ball? forward submitprevious answers correct
Using the equation of part a, we can include in it the directions:
v1fx = ((m1-m2) / (m1 + m2)) v1ix
v1i: initial velocity of ball 1 in the direction of the x-axis
v1f: final speed of ball 1 in the direction of the x-axis
like m1 >> m2 then
v1fx = v1ix
v1fx = 200 m / s (positive x direction)
So it is concluded that the ball 1 continues forward.
answer:
forward
part c part complete what is the final speed of the stationary ball if the incoming ball is much more massive than the stationary ball ?.
The shock is perfectly elastic. For this case, we have that the equation to find the final velocity of the stationary ball is given by
v2f = ((2m1) / (m1 + m2)) v1i
where,
v1i: initial speed of ball 1.
v2f: final speed of ball 2.
m1: mass of the ball 1
m2: mass of the ball 2
Then, as we know that m1 >> m2 then
v2f = ((2m1) / (m1) v1i
v2f = 2 * v1i
v2f = 2 * (200 m / s)
v2f = 400 m / s
answer
400m / s
