# Physics

sorry i know i keep asking but my teacher wont be here today and tomorrow..and i don't understand these four problems...

1) a 65.0-kg person throws a .045kg snowball forward with a ground speed of 30.0 m/s. a second person with a mass of 60.0-kg catches the snowball. both people are on skates. the first person is intially moving forward wit ha speed of 2.50 m/s and the second person is initially at rest. what are the velocities of the two people after the snowball is exchanged? disregard the friction between the skates and the ice.

2. a 7.00 kg bowling ball collides head-on with a 2.00-kg bowling pin that was originally at rest. the pin flies forward with a speed of 3.00 m/s. if the ball continues forward with a speed of 1.80 m/s, what was the initial speed of the ball? ignore rotation of the ball

3. an 8.00g bullet is fired into a 250-g block that is intially at rest at the edge of a table of 1.00 m height. the bullet remains in the block and after the impact the block lands 2.00 m from the bottom of the table. determine the initial speed of the bullet.

4. a billard ball moving at 5.00 m/s strikes a stationary ball of the same mass. after the collision, the first ball moves at 4.33 m/s at an angle of 30 with respect to the original line of motion A) find the velocity (magnitude and direction) of the second ball after the collision. B) was this an inelastic collision or an elastic collision?

err >> please someone or ppl help..i really don't understand this

thanks

Please post your questions one at a time; that way they will be answered sooner and probably in more detail.

2. Whether or not the collision is elastic, momentum must be conserved. So just set the inital and final total momenta of both balls (combined) equal to each other. In this case, the only unknown will be the ball's initial velocity. Letting capital M and V represent the ball and lower case m and v represent the pin,
M V1 = M V2 + m v2
7 V1 = 7 * 1.8 + 2*3 = 18.6 kg m/s
V1 = ?

ohhh okay sorry about that XD

shall i just do it that next time i might have problems?

From the fact that the block lands 2 m from the table, and the height H, you can calculate the velocity V2 of the block after the bullet goes inside it.
V2 = (2 meters)/T

where T is the time required for the block to hit the ground.
(1/2) g T^2 = H
T = sqrt (2H/g)
V2 = (2 meters)/sqrt (2H/g)= 4.43 m/s

Now that you know V2, use conservation of momentum to get the initial velocity of the bullet. This part is very similar to problem 2.
V1 m = (M + m) V2
V1 = (258/8) * 4.43 m/s = __ m/s

ok so what i would do is first off convert (258) to kgs and then

doing (.258kg)(4.43m/s)=1.14 m/s?

No. Please reread my answer. 4.43 m/s gets multiplied by the RATIO 258g/8g = 32.25. You do not have to convert the grams to kg before calculating the ratio, because the ratio is dimensionless.

In case it was not made clear, the 258/8 ratio is (M+m)/m = (250 +8)/8

This is another momentum conservation problem. The initial momentum of everythng (person 1, person 2,and the snowball) is
(M1+m) V1i, where M1 = 65 kg and V1i = 2.5 m/s, and m = 0.45 kg is the mass of the snowball that person 1 initially carries. If you consider the momentum of M1 and the snowball when it is thrown,
(M1 + m)V1i = M1 V1f + m v
where v = snowball ground speed of 30 m/s. This lets you solve for the final velocity of person 1, V1f
V1f = [M1+m)/M1]V1i - (m/M1)v

To get the final velocity of person 2,V2f, consider momentum conservation before and after the snowball hits him.
m v = (M2 + m) V2f
Solve for V2f = [m/(M2+m)] v.

why is it divided by 8? because don't you need to make it into kg conversion first?

oh wait sorry i didn't read the message before that..

i think i get this one now

umm for the 4th one i got the direction because it would be perpendicular if i am correct? so it would be -60 degrees. and then it is a elastic collision. i found out the solution for the magnitude is supposed to be 2.5 m/s but im not getting that. so i dunno if i am doing something wrong. but in addition i don't understand cause it says that there are first and second ball and im unclear about that question by itself too

Billiard ball collsions are very close to being elastic, but all we can assume to start this problem is that MOMENTUM is conserved (in any direction), and then see if the answer tells us if the collision was elastic or not.

<<a billard ball moving at 5.00 m/s strikes a stationary ball of the same mass. after the collision, the first ball moves at 4.33 m/s at an angle of 30 with respect to the original line of motion A) find the velocity (magnitude and direction) of the second ball after the collision. B) was this an inelastic collision or an elastic collision? >>

Let m be the mass of either ball. Write two equations of momentum conservation, in directions along and perpendicular to the first ball's initial direction. Let A be the agle of the "hit" ball with respect to the original direction (x axis). Let v be the speed of the second ball after collision, and V = 5.00 m/s be the initial velocity of the first ball, and V' = 4.33 m/s be the velocity of the first ball.
M V = M V' cos 30 + M v cos A
0 = M V' sin 30 + M v sin A
Cancel out the M's and plug in the numbers you know

v cos A = 5 - 4.33 cos 30 = 1.25 m/s
4.33 * sin 30 = - v sin A
v sin A = 2.165 m/s
v^2 = 2.165^2 + 1.25^2 = 6.25 m/s
v = 2.5 m/s
You can get the angle A from the sin A/cos A ratio.
V^2 = 25 m^2/s^2 (before collision)
V'^2 + v^2 = 18.75 m^2 + 6.2562 = 25 m^2/s^2 (after collision)
The means the sum of the kinetic energies stayed the same. The collision was therefore elastic.

If you had known in advance that the collision would be elastic, you would only need one "known" (for example, the direction of one ball) to solve for all other variables.

There are some typo mistakes in my last post. M and m are the same. The last equatiuon should read
V'^2 + v^2 = 18.75 m^2/s^2 + 6.25 m^2/s^2 = 25 m^2/s^2 (after collision)

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1. if ur on about somthing else sorry but there are 1000grams im a killogram

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posted by lauren

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