A ball of mass 100 g travelling with a velocity of 100 m/s collides with another ball of mass 400 g moving at 50 m/s in the same direction. If they stick together, calculate:
(i) their common velocity
(ii) loss of enery and hence state the type collision involved.
I had calculated their common velocity of which i got 60m/s.
Please help me (ii)
linear momentum p=mv
lets say m1 =0.1 kg
m2=0.4kg
before collision pi=(m1. u1)+(m2 . u2)
pi=(0.1 . 100) + ( 0.4.50) = 30
when they collide together pf= (m1+m2)vf
since linear momentum is always conserved pi=pf
pf= (m1+m2)vf =pf
(0.1+0.4) vf = 30
vf=60m/s
you are correct
b)
when two objects stick together their collision is a PERFECTLY inelastic collision
you now have to calculate ki and kf
ki-kf=loss of energy
kf = 1/2 . (m1+m2). vf^2
kf = 1/2 . (0.1+0.4). 60^2=900j
ki = 1/2 .m1 . u1^2 + 1/2 m2 . u2^2
ki = (1/2 . 0.1. 100^2 ) + (1/2 . 0.4 . 50^2)=1000J
energy lost = 100 J
ki is the kinetic energy initial how much energy is there in the system in the form a speed at the beggining of the collision as you know there are two forms of mechanical energy ( kinetic energy and potential ) when the object is moving it has kinetic energy but when it collides some of the kinetic energy transforms into heat and sound thus the speed of the object slows down.When we calculate the Kf we calculate the energy after the collision.Like I said during an Inelastic collision kf is less then ki because there has been lost of energy ( transform into heat and sounds etcc
Meaning of pi and pf
wow! it's great! Thank you somuch AMELIA.
But please help me with the meaning of ki and kf
Thanks
for the solving
SswPlease i dont really understand this answer
I dont understand
Thanks for that
To calculate the loss of energy and determine the type of collision involved, we need to use the principle of conservation of kinetic energy. In an isolated system (where no external forces act), the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Let's first calculate the initial kinetic energy of each ball:
Kinetic energy of the first ball (m1) = (1/2) * mass1 * (velocity1)^2
= (1/2) * 0.1 kg * (100 m/s)^2
= 500 J
Kinetic energy of the second ball (m2) = (1/2) * mass2 * (velocity2)^2
= (1/2) * 0.4 kg * (50 m/s)^2
= 500 J
The total initial kinetic energy before the collision is the sum of the kinetic energies of the individual balls:
Initial total kinetic energy (E_initial) = (500 J) + (500 J)
= 1000 J
Now, let's calculate the final kinetic energy after the collision. Since the balls stick together, they will move with a common velocity (v) after the collision:
Final kinetic energy (E_final) = (1/2) * (m1 + m2) * (v^2)
Substituting the values:
1000 J = (1/2) * (0.1 kg + 0.4 kg) * (v^2)
2000 J = 0.5 kg * v^2
v^2 = (2 * 2000 J) / 0.5 kg
v^2 = 8000 J/kg
The common velocity (v) is the square root of 8000 m^2/s^2, which is approximately 89.44 m/s.
Now, let's calculate the loss of energy (ΔE) in the collision:
ΔE = E_initial - E_final
= 1000 J - 8000 J
= -7000 J
The negative value indicates a loss of energy, which implies that the collision is inelastic. In an inelastic collision, kinetic energy is not conserved as it transforms into other forms of energy (such as heat or sound) during the collision.
Therefore, the common velocity after the collision is approximately 89.44 m/s, and the collision is inelastic with a loss of 7000 J of energy.