Steel Ball Collision: Calculating Velocity and Direction

AI Thread Summary
A 2.0 kg steel ball moving at 5.0 m/s [W] collides with a stationary ball of equal mass, resulting in the first ball being deflected [N35W] at 3.0 m/s. The total momentum of both balls before the collision is zero in the [NS] axis, indicating it must also be zero after the collision. The angle between the velocities of the two balls post-collision is 90 degrees due to their equal masses. To solve for the second ball's velocity, one must apply the conservation of momentum equations for both the [NS] and [WE] axes. The original poster eventually resolved their confusion after realizing a calculation error.
3.14159265358979
Messages
43
Reaction score
0
1) A 2.0 kg steel ball rolling at 5.0 m/s [W] strikes a second steel ball of equal mass at rest. After a the collison, the first ball is deflected [N35W] at 3.0 m/s. Determine the velocity (and direction) of the second ball. (the answer in the back of the book is 4.1 m/s [s37w], but i got something different) thanks!
 
Physics news on Phys.org
What is the total momentum of the two balls before the collision, in the [NS] axis? Zero. So what should be the total momentum of the two balls after the collision along the same axis? Zero as well. So you have one equation, for the conservation of momentum in the [NS] axis, with one unknown - the velocity of the second ball after the collision. (In case this is what you are missing, the angle between the velocities of the two balls after the collision is 90 degreesm, because the masses of the balls are equal. If you cannot use this piece of information, or need to prove it first, just write down the equation for the conversation of momentum in the second axis, [WE], and introduce another variable into your equations - the angle of the velocity of the second ball.)
 
Last edited:
i still don't understand how they got that answer! help please? thank you..
 
You say you got something different? Let's see what you did! That way we're not all reinventing the wheel.

cookiemonster
 
nevermind...i finally got it...i was making the most stupid mistake..
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top