Abeka Physics Section 12.2 pg. 178 - 183 Flashcards
Define momentum
> momentum of an object is equal to the product of its mass and velocity
p=mv where p stands for momentum
Solve this: A cannonball has a mass of 18 kg. If it is moving along a line 20degrees north of east at a speed of 200. m/s, what is its momentum?
> use equation p=mv, where m is 18kg and v is 200.m/s.
p=mv
p=(18kg)(200.m/s)
p=3600 kg x m/s in the direction of 20 degrees north of east
Explain how F = delta p / t is derived
> from Newton’s 2nd law of motion, F = ma
a = v(sub f) - v(sub i) / t
so, F = m x delta v / t
according to momentum equation, p=mv,
delta p = p(sub f) - p(sub i) = mv - mv
assuming mass does not change m(sub f) = m(sub i), this simplifies to delta p = m x delta v
substituting delta p for m x delta v in equation p=mv gives
F = delta p / t
State and describe Newton’s original statement of his second law of motion
> “force applied to object equals time rate of change in momentum”
“greater the force applied to object, more quickly its momentum will change”
F = delta p / t
Solve this: If the fly can muster 2.00 x 10^-7 N of force against a locomotive with a mass of 6.40 x 10^5 kg traveling at a constant 22.0 m/s, how long will the fly have to continue pushing before he stops the locomotive? Assume that the forces producing uniform motion in the locomotive before the fly intervenes do not change.
> find the time from equation F= delta p / t, where delta p is due to a change in velocity, not a change in mass.
substitute equation delta p = m x delta v into equation F = delta p / t and solve for t to obtain
F = delta p / t = m x delta v / t
t = m x delta v / F
since final velocity is 0 m/s, final momentum of locomotive will also be zero. Therefore, t = m x -v(sub i) / F =
-mv(sub i)/ F
since the fly is pushing opposite of the train’s motion, F is -2.00 x 10^-7 N
substituting this and the other information gives
t = -(6.40x10^6kg)(22.0m/s) / -(2.00x10^-7N)
t = 7.04 x 10^13 s
since there are about 3.15 x 10^7 s in a year, it takes the fly only 2.23 x 10^6 years. He can do it; he just needs to be patient.
Describe one of the 2 useful interpretations of a Newton’s second law in its original form
> newton’s second law of motion shows how much force must be continually applied during given interval to accelerate object from rest v(sub i) = 0 to velocity v(sub f)
then equation F = m x delta v / t becomes
F = m x v(sub f - 0) / t = mv(sub f)/t
is evident that if 2 objects are to be accelerated to same velocity in same time, required force depends on their masses
Describe 2nd useful interpretation of Newton’s second law in its original form
> law shows how much force must be continuously applied during given interval to stop v(sub f)= 0 object initially moving at velocity v(sub f)
F = m( 0 - v(sub i) / t = - mv(sub i)/t
negative sign indicates that force is opposite to motion
solve this: How much force continuously applied for 3.00s must be used to stop a hockey player (882 N) sliding at an initial velocity of 7.5 m/s on frictionless ice?
> time t is 3.00s
since hockey player eventually comes to stop, v(sub f) = 0
initial velocity, v(sub i) is 7.5 m/s
to find m use equation m = F(sub w)/g where F(sub w) is 882N and g is 9.80m/s^2
m = 882N/9.80m/s^2
m=90.0kg
finally substitute into equation F=m x delta v /t
F= (90.0kg)(0 - 7.50m/s^2) / 3.00s
F = -225 N (about -50.6 lb)
What does the law of conservation of momentum state?
> law of conservation of momentum: “when the net force acting on a system is zero, the momentum of the whole system remains constant”
What does the law of conservation of momentum state?
> law of conservation of momentum: “when the net force acting on a system is zero, the momentum of the whole system remains constant”
Describe the law of conservation of momentum
> newton’s first law states that any material object has inertia; that is, in absence of external force, object will maintain its present velocity
another way of expressing same idea is to say that object’s velocity is conserved if no net force acts on object
isolated system of objects also has inertia, but inertia does not conserve velocity in system of objects
rather, conserved quantity is momentum of whole system; thus, while velocities of individual objects may change, total momentum of system does not
as long as system is free of any net external force, momentum is conserved
Define momentum of a system of objects
> momentum of a system of objects - sum of their individual momenta
using the law of conservation of momentum, what could a sports player predict?
> using the law of conservation of momentum, a sports player could predict the exact outcome of each collision
Describe colliding objects in regards to the law of conservation of momentum
> imagine baseball and tennis ball rolling toward each other along floor
before they collide, baseball (object A) has momentum of m(sub A)v(sub A) and tennis ball (object B) has momentum of m(sub B)v(sub B)
after they collide, both balls have new velocities, which we will call v’(sub A) and v’(sub B)
therefore the new momentum of baseball is
m(sub A)v’(sub A) and the new momentum of tennis ball is m(sub B)v’(sub B)
since total momentum of system remains constant,
m(sub A)v(sub A) + m(sub B)v(sub B) = m(sub A)v’(sub A) + m(sub B)v’(sub B)
this equation must be satisfied in any collision between two objects
solve this: A 40.0g marble rolling at 0.550 m/s collides with another 40.0g marble sitting at rest. If the rolling marble comes to a dead stop after the collision, what velocity is imparted to the resting marble? Assume no force of friction
> if rolling marble is A and resting marble is B, v(sub a) and v’(sub A) are both 0 m/s
thus, equation m(sub A)v(sub A) + m(sub B)v(sub B) = m(sub A)v’(sub A) + m(sub B)v’(sub B) becomes
m(sub A)v(sub A) = m(sub B)v’(sub B)
since m(sub A) is 40.0g, v(sub A) is 0.550m/s, m(sub B) is 40.0 g, and the unknown is v’(sub B), solve for v’(sub B) to obtain
v’(sub B) = m(sub A)v(sub A)/ m(sub B)
Thus, v’(sub B) = 40.0g(0.550m/s)/40.0g
v’(sub B) = 0.550m/s
resting marble shoots away at a speed equal to that of the rolling marble
