Chapter 7 - Impulse & Momentum Flashcards
A 46-kg skater is standing still in front of a wall. By pushing against the wall she propels herself backward with a velocity of −1.2 m/s. Her hands are in contact with the wall for 0.80 s. Ignore friction and wind resistance. Find the magnitude and direction of the average force she exerts on the wall (which has the same magnitude as, but opposite direction to, the force that the wall applies to her).
Question is: what force are you calculating when you calculate the momentum?
Impulse is the change in momentum and the force is what caused that change.
In this example, the force of the wall caused that change
Which external force caused her change in motion? The force of the wall
So when you solve for her change in momentum you get initial is 0 and final is -55
dividing by time to get the force you know that you have to change signs so it’s +69
Two skaters collide (in a 2D collision) and
embrace, in a completely
inelastic collision. That is, they
stick together after impact, as
shown by figure, where the
origin is placed at the point of
collision. Alfred, whose mass
mA is 83 kg, is originally moving
east with speed vA
= 1.7 m/s.
Barbara, whose mass mB is
55 kg, is originally moving north
with speed vB
= 2.1 m/s. What is
the change in the total kinetic
energy of the skaters because of
the collision?
1) You calculate total initial KE which is easy, sum of Alfred and Barbara’s initial kinetic energies
2) To find the final kinetic energy you need the final speed
3) You can do this by finding momentum, which is conserved in an inelastic collision. Find the inital momentum in the x and y directions, and set this equal to the final momentum in the x and y directions
4) This will allow you to isolate for vfx and vfy and use the Pythagorean theorem to find the vf value, you should get 1.32
5) Sub this vf value into final kinetic energy and subtract from initial to figure out change in total kinetic energy
you should get final ke is 120
change in je is -120
In Figure, block 1 of mass m1 slides from rest along a frictionless ramp from height h =
2.50 m and then collides with stationary block 2, which has mass m2 = 2.00 m1. After
the collision, block 2 slides into a region where the coefficient of kinetic friction µk is
0.50 and comes to a stop in distance d within that region. What is the value of distance
d if the collision is (a) elastic and (b) completely inelastic?
Hint: For elastic collision with stationary target (v2i = 0): v2f = 2miv1i/m1 + m2
1) First you want to find the speed of block 1 right before the collision. Use kinematic equations to do so and obtain 7
2) Now you want to find the velocity of each block after the collision. To do so for elastic, use equation provided.
You should get vf = 4.67
Obtain equation from inelastic using the m1vf + m2vf = m1v0 + m2v0 formula, knowing that m2vo = 0 and there is a combined vf.
You should get vf = 2.33
3) Use the work energy tehorem which states taht work is equal to the change in kinetic energy.
Work is frictional force which is fk = usmg and then work done by firction is usmgd = 1/2m2v2^2 (no inital kinetic energy of block 2 becuase of no speed
For the elastic condition you should get d = 2.22
For the inelastic condition yuou should get d = 0.53
In Figure, puck 1 of mass m1
= 0.20 kg is sent sliding across a
frictionless lab bench, to undergo a one-dimensional elastic collision
with stationary puck 2. Puck 2 then slides off the bench and lands a
distance d from the base of the bench. Puck 1 rebounds from the
collision and slides off the opposite edge of the bench, landing a distance
2d from the base of the bench. What is the mass of puck 2? (Hint: Be
careful with the signs.)
1) You want to setup the equations
x = vt, subbing in the value for vf that is listed below
Then you want to set up a ratio and do x1/x2 = -2d/d (make sure the -2 is negative!!!)
velocity equations are given
and then sub in teh values and you should be left with only m1 and m2, be able to isolate for m2 and sub in teh m1 value!
you should get 0.4kg
Equation for impulse
J = netforceT
esentially the change in momentum
∆P = mvf - mv0
In what scenario can you apply the conservation of momentum?
When the system is isolated, in other words the net external force is 0
Conservation of momentum and kinetic energy equations
Momentum
m1vf1 + m2vf2 = m1v01 + m2v02
Kinetic energy
1/2m1vf1^2 + 1/2m2vf^2 = 1/2m1v01^2 + m2v02^2
In elastic and inelastic collision is momentum and kinetic energy conserved?
In elastic collision, kinetic energy and momentum is conserved
In inelastic collision, momentum is conserved but not kinetic energy
In completely inelastic collision, objects stick together so they have the same final velocity
Center of mass equation adn velocity of center of mass equation
Center of Mass
Xcm = (m1x1 + m2x2)/m1 + m2
Velocity of center of mass
Vcm = (m1v1 + m2v2)/m1 + m2
