In simplest physics terms (not accounting for friction or minor loss) assuming total energy transfer... a 100 gram ball moving at 4 mph striking a stationary ball 25 grams.
Would the small ball move at 16mph after collision? (4x faster becasue 1/4 mass?)
or
would it move at the maximum 4mph and the added energy would remain in the larger ball?
Thank you all.
electron1
If all of the kinetic energy of the 100 gram ball is transferred to the 25 gram ball during the collision, the 100 gram ball’s velocity after the collision would be 0 m/s. To determine the velocity of the 25 gram ball after the collision, we need to determine the initial kinetic energy of the 100 gram ball.
KE = ½ * m * v^2
One mph = 0.447 m/s
v = 4 * 0.447 = 1.788
KE = ½ * 0.1 * 1.788^2 = 0.1598472
½ * 0.025 * v^2 = 0.1598472
v^2 = 0.1598472 ÷ 0.0125= 12.78776
v = 3.576 m/s
This is not 4 times the velocity of the 100 gram ball.
3.756 2 ÷1.788 = 2
This is 2 times the velocity of the 100 gram ball. This is because kinetic energy is directly proportional to the square of the velocity.
Initial momentum = 0.1 * 1.788 = 0.1788
Final momentum = 0.025 * v
0.025 * v = 0.1788
v = 0.1788 ÷ 0.025 = 7.152
7.152 ÷ 1.788 = 4
If we use conservation of momentum, the velocity of the 25 gram ball is 4 times the initial velocity of the 100 gram ball. Let’s determine the kinetic energies.
For the 100 gram ball, KE = ½ * 0.1 * 1.788^2 = 0.1598472 J
For the 25 gram ball, KE = ½ * 0.025 * 7.152^2 = 0.6393888 J
This is impossible. The kinetic energy of the 25 gram ball can’t be greater than the kinetic energy of the 100 gram ball. This means the 100 gram ball must have kept on moving after the collision. The only way to solve this problem is to assume the collision is elastic. In this type of collision, kinetic energy is conserved. This means only part of the 100 gram ball’s kinetic energy will be transferred to the 25 gram ball.
Steve
If you assume total energy transfer, you will violate conservation of momentum.
If, instead, you assume NO ENERGY LOSS, then this is an elastic collision, and
Vcm = P/Σm = 100*4/(100+25) = 3.2 mph
V25 = Vcm + Vcm = 6.4 mph and
V100 = Vcm - (4 - 3.2) = 2.4 mph
Check: 6.4*25 = 160;.....(4 - 2.4)*100 = 160
Each ball has the same dP
David N
The key words are: ASSUMING TOTAL ENERGY TRANSFER.
If all of the energy is transferred to the small ball, then the large ball would be motionless.
With conservation of momentum:
M(i-large) = 100 g * 4 mph = 400 g*mph
M(i-small) = 25 g * 0 mph = 0 g*mph
M(i) = 400 g*mph + 0 g*mph = 400 g*mph
M(f) = 100 g * 0 mph + 25 g * x mph = 0 + 25 g * x mph = 400 g*mph
25x = 400
x = 16
Therefore v(f-small) = 16 mph.
oubaas
which sort of collision ?
if anelastic, then just momentum is conserved
0.1*4 = (0.1+0.025)*V..they both proceed with same versus and speed
if elastic, then also energy is conserved
0.05*4^2 = 0.05*Va^2+0.0125Vb^2
0.1*4 = 0.1*Va+0.025Vb
solving this twin equation system we get Vb = 6.40 mph and Va = 2.40 mph
they both proceed with same versus and different speeds
The momentum variation ΔI experienced by the bigger ball ((4-2.4)*100 = 160 g*mph) is exactly equal to the momentum variation experienced by the smaller ball (6.4*25 = 160 g*mph)
The energy variation ΔI experienced by the bigger ball ((4^2-2.4^2)*50 = 512 g*(mph)^2) is exactly equal to the energy variation experienced by the smaller ball (6.4^2*12.5 = 512 g*(mph)^2)