Abeka Physics Section 12.3 pg. 183 - 184 Flashcards
Define and describe impulse
> impulse - any change in momentum
impulse (J) = Ft [product of force causing change and length of time force is applied]
is vector in same direction as applied force
measured in N x s
Describe what happens when a force is applied by a hockey stick on a puck
> impulse applied to object creates change in velocity when it changes object’s momentum
impulse is equal to resulting change in momentum as puck increases in velocity
impulse is not property of object, but is rather measure of extent to which object is affected by force
Describe how impulse is shown on a graph
> force applied to object is on y-axis
time force is applied to object is on x-axis
area underneath curve created by force and time values is the impulse
Explain why a long, sweeping kick make a soccer ball travel farther than a short, quick kick.
> longer constant force is applied to object of constant mass, greater velocity change
full stroke increases time of contact, thereby increasing impulse and velocity change
Solve this: A 1.0kg ball when dropped hits the floor at a speed of 25m/s. It rebounds with a speed of 9.0m/s. (a) What impulse acts on the ball during contact with the floor? (b) If the ball is in contact with the floor for 0.020s, what is the average force exerted on the floor?
(a) Use equation J = m[v(sub f) - v(sub i)]
>it is stated that m is 1.0kg
>since ball falls downward and rebounds upward, v(sub i) is -25m/s, and v(sub f) is 9.0m/s.
>substitute to obtain
J = (1.0kg)[9.0m/s - (-25m/s)]
J = 34 kg x m/s
>this means that the impulse is delivered upward, opposite the initial motion
(b) Solve J=Ft for F; then replace J with the impulse computed in (a) and t with 0.020s:
F = 34 kg x m/s / 0.020s
F = 1700 N or about 380 lb.
>this is average force floor exerted on the ball. On a side note, the ball’s average force exerted on the floor is equal and opposite: -1700 N
