How many collisions will take place?
112172
Only a limited answer:  For the special case where M = m, there are just 2 collisions.  After the first collision, V = 0 and v = V's initial value.  In the post-reflection collision the opposite exchange occurs.
91362
It's tranferred to heat/friction from the impact
816105
Generally, some of their collective kinetic energy is converted into heat, so the KE after the collision is less than the initial KE.  Momentum is conserved.
866141
You have to calculate p=massXspeed before the "collision".  The quantity remains constant, so if the mass increases, the speed must decreases by the same ratio for the p to stay the same.  (First, you have to convert the 8N weight to a mass using the 9,8N/kg gravity constant)
1081829
Elastic collision.  An elastic collision is a collision in which the total kinetic energy of the colliding bodies after collision is equal to their total kinetic energy before collision. Elastic collisions occur only if there is no conversion of kinetic energy into other forms, as in the collision of atoms.  In this case, two bodies (the ball and the earth) collide.  Since  the mass of the ball is very small compared to the mass of the earth and kinetic energy must be conserved, the ball bounces back with a velocity.
1053533
Depends on how many times he gets hit in the head.
154434
The problem cannot be attended to without knowing the total number of possibilities.\n\nhow many sections are there on the spinnner..\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/
253229
If you mean 'how many will be present' then the answer is 736. If you mean 'how many will play' then nobody knows. It depends on injuries, suspensions and squad rotation. I would expect somewhere between 540-550 will actually play at some point.
788625
This is a momentum question.  If the end result is comming to a halt, then then each vehicles' momentum was equal and opposite.  Momentun (M) = mass (m) * velocity (v),  therefore you are left needing to solve the equation m1*v1 = m2*v2...you can do the rest I hope....
19818
try\nhttp://www.peterwebb.co.uk/probability.htm
328332
That depends on where the incident occurs.
926868
Solve for momentum.\n\nBefore collision total momentum  = mv\n\nDefine your coordinate system so that the particle was traveling along the x axis in the positive direction, and the collision happens at the origin. Then all of the momentum before the collision is in the x direction. the initial y component of momentum is zero.\n\nparticle 1 goes flying off at v1 at angle a1\n\nparticle 2 goes flying off at v2 at angle a2\n\nThe total momentum afterwards must be equal to the momentum before. So the y component of p1 plus the y component of p2 must equal the y component of the momentum before collision (zero).\n\nSo m*v1*sin(a1) = m*v2*sin(a2)\n\nSimilarly the x components must add up so \n\nm*v1*cos(a1) + m*v2*cos(a2) = mv\n\nNow just solve the first equation for V1 and substitute in the second equation. \n\nThen solve the second equation with substitutions to find the value of v2 in terms of v, and the sin and cos of the two angles. \n\nFinally substitute the value of v2 back into your equation for v1 to find the value of v1.\n\nI leave the simple algebra of solving the equations and the calculation of plugging the numbers into the equations to the student.
349995
It has 3 speeds, stop, go and crash.
15239
The correct answer is 190. It would involve any spare (all pins down on two balls) in the first frame, then single-pin spare conversions in frames 2-9 and a 9-count, spare and another 9-count in the tenth and final frame.
365970
1/161\nTake the number of ways the event can occur divided by the total number of events.
1065918
To get your answers, imagine what has to happen.  \n\na) First, you have to pick 1 of the 4 white balls out of the 12 balls there.\n\n2nd, you have to pick 1 of the 3 white balls left out of the 11 balls there.\n\nSo, in order to get a red and then a red, you need to get 4 out of 12 followed by 3 out of 11 or 4/12 times 3/11 or 12/132 = 1/22.\n\nb) First, how many ways can 5 people line up?  5!.  How many ways are correct (assuming they all have different names, only 1).  So, the prob is 1/5! or 1/120.
947533
Question is how many will we stop?