There are two balls of same mass,same material......?
336191
Well, if they are the same mass and the same material, and you KNOW this before hand, it would be obvious.  \n\nThe larger ball would be the hollow one because being the same mass they would have to weigh the same.  A hollow ball would have to be bigger to match the mass of the solid ball.
878067
As long as neither of the balls starts spinning (or changes the rate at which it's already spinning) after the collision (in which case some of their linear momentum would have been converted to angular momentum), the total linear momentum of the system before and after the collision must be the same.\n\nLet p11 and p21 be the momentum of balls 1 and 2 before the collision; let p12 and p22 be the momenum of balls 1 and 2 after ths collision; and let m be the mass of the balls (which we are told have the same mass).  The momenta are all vector quantities, which I'll write as order pairs of numbers (i.e.,  p = (x,y) = x*i + y*j, where i and j are the unit vectors in the +x and +y directions, respectively).\n\nLet the +x axis be pointed in the east direction, and the +y axis be in the north direction.\n\nThen, writing each \n\np11 = m*(2 m/s, 0 m/s)\np21 = m*(-1.2 m/s, 0 m/s)\np22 = m*(0 m/s, 1.43 m/s)\np12 = m*v12 (v12 is the vector velocity of ball 1 after the collision)\n\nBy conservation of momentum, we have that:\n\np11 + p21 = p12 + p22\n\nm*(2 m/s, 0 m/s) + m*(-1.2 m/s, 0 m/s) = m*v12 + m*(0 m/s, 1.43 m/s)\n\nv12 = ((2 - 1.2 - 0) m/s, (0 + 0 -1.43) m/s)\n\nv12 = (0.8 m/s, -1.43 m/s)\n\nThe speed of ball 1 after the collision is given by:\n\nsqrt((0.8 m/s)^2 + (-1.43 m/s)^2) = 1.639 m/s\n\nThe angle the velocity vector of ball 1 makes with respect to the +x axis after the collision is given by:\n\narctan(-1.43/0.8) = -1.061 radians = -60.776 degrees\n\n(It helps to draw a picture of this to visualize why this relationship holds.)\n\nThe first ball is traveling 60.776 degrees south of east after the collision.
38813
density has an equation density=mass/volume\nClay in a ball has the same mass and volume as rolled into a ball (considering of course no clay comes off on the table). So the density would be calculated the same,no matter the shape.
495603
Assuming that the energy of the ball remains constant (i.e. no air resistance), the speed will be the same. The VELOCITY, however, will be -35 m/s.\nSpeed and velocity are NOT the same!
84559
Both marbles will hit the ground at the same time, but one will roll along farther.\n\nIn short, the vertical force from gravity is the same for both marbles.  The one hit sideways just has an additional horizontal velocity in addition to its vertical acceleration.
312163
the first after collision moves in an opposite direction and the second ball moves towards the direction of collision\nfirst ball moves with a velocity of  16.66cm/s\nsecond ball with 66.66cm/s
1171138
They have the same atomic number but different mass numbers?\nOdd kind of an insult...
262517
They are the same person.
299265
The answer has to be A.\nThe time will be the same because Gravitational force does not depend on the mass of the object, since they start with same velocities and accelerate with the same rate they must hit the ground at the same time.\n\nAnd it obviously has to be the same distance because if it wasn't then they would never hit the ground at the same time in the first place. That would happen because if 2 bodies start with the same speed and have the same acceleration but different displacements then their time has to be different.
639089
They are one in the same.
711513
Hi!  You are right.\nIf they have equal masses and are elastic then  b) it is going to happen. Why? well Vf= 2*Vcm-Vi. Where all are vectors and :\nVf= final speed;   Vcm=center of mass speed ;  Vi=initial speed.\nThus 0=2*Vcm-Vi => Vcm=V/2=(m1V1+m2V2)/(m1+m2)=\n=m1*Vi/(m1+m2)  => m1=m2.    \n  " a)" isn't possible because the total momentum must be preserved.\n          Hope it helps!
894241
answer is approxomately 42.476m/s\n\nonly one formula here \nS=ut + (at^2)/2 \ns- distance travelled = 40m\nu- initial valocity\na - acc due to gravity-9.8m/s^2\nt- time \nin the first case the initial velocity is 0m/s^2\nso\n40=(9.8*t^2)/2\nt=2.857 sec\nas both the balls have to touch the ground at same time  and the second ball being dropped( rather thrown would sound correct) after 2 secs, the later has only 0.857 sec left for traelling a distance of 40m. \nso \n40=u*0.857+(9.8*(0.857^2))/2\nsolving u=42.476m/s\nthe initial velocity for the second ball should be 42.476m/s and should be thrown vertically downwards.\nthat is pheta=0 in the formula given un the question.
759986
both will have same weight mass
171753
oldhippypaul is right, the correct statement is "both corks are subject to the same buoyant force". The explanation is that, if they´re identical,they weigh the same. Since there in balance (i.e., there not sinking or emerging) the bouyant force must equal thre weight and, therefore is fhe same.
58960
This is absolutely true.  My high school basketball coach demonstrated this to us when the team was struggling with making free throws.  I wish I had a picture to show you to confirm this, but you'll have to take my word for it.\n\nNote:  This is true with a regular sized ball.  Actually, if you're good at math, you can prove this by measuring the diameter of a regulation sized rim and comparing it against the diameter of a basketball :)  \n\nLastly, there is one URL which also states similar belief that 2 basketballs can fit in the rim at the same time.  See source below.\n\nHope this helps!  :)
936185
They have the same mass. An electron essentially has no mass. so both have the same mass.
904552
Both balls would hit the rocks at the same speed.\nBall #1 would leave your hand travelling straight up at speed 'u'. It travels up, stops, and starts plummeting back to earth. When it reaches the height at which it left your hand, it is travelling at speed 'u' straight down. The speed that it hits the rocks at is u+(time x 10m/s/s).\n\nBall #2 would leave your hand travelling straight down at speed 'u'. The speed that it hits the rocks at is u+(time x 10m/s/s). It is the same speed.