A 2.20-kg ball, moving to the right at a velocity of +1.27 m/s on a frictionless table, collides head-on with a stationary 7.20-kg ball. Find the final velocities of (a) the 2.20-kg ball and of (b) the 7.20-kg ball if the collision is elastic. (c) Find the magnitude and direction of the final velocity of the two balls if the collision is completely inelastic.
I solved part b, I need help with part a if possible, cause if I solve a I could easily solve part c
sorry I solved part C, I need help with a and b pls.
ok I got the answer:
a) Vf1=(m1-m2/m1+m2)*(Vi1)
B) Vf2=[(2(m1)/m1+m2](Vi1)
Final Velocity of 2.2kg ball:
Vf = (V_i *(m1-m2) + (2m2*v_i2)/(m1+m2)
v_i =1.27
m1=2.20
m2=7.2
v_i2 = 0 (because it was a rest initially)
Vf=(1.27*(2.2-7.2)+(2*7.2*0)/(1.27+7.2))
=-6.35
For final Velocity of 2nd object:
Vf = (V_i2 *(m2-m1) + (2m1*v_i/(m1+m2)
Vf=(0*(7.2-2.2)+(2*2.2*1.27)/(1.27+7.2))
Vf=.66
So answer is
-6.35 and .66
Someone check my method and math.
yeah you are right, ignore my post.
To solve part (a) of the problem, we can use the principle of conservation of momentum. In an elastic collision, both momentum and kinetic energy are conserved.
Let's denote the mass of the 2.20-kg ball as m1 and the mass of the 7.20-kg ball as m2. The initial velocity of m1 is +1.27 m/s, and the initial velocity of m2 is 0 m/s since it is stationary.
According to the law of conservation of momentum:
Initial momentum = Final momentum
Initial momentum of m1 + Initial momentum of m2 = Final momentum of m1 + Final momentum of m2
m1 * v1 initial + m2 * v2 initial = m1 * v1 final + m2 * v2 final
Now, let's plug in the given values:
m1 = 2.20 kg
m2 = 7.20 kg
v1 initial = +1.27 m/s
v2 initial = 0 m/s
The final velocities are unknown, so let's denote them as v1 final and v2 final.
The equation becomes:
2.20 kg * 1.27 m/s + 7.20 kg * 0 m/s = 2.20 kg * v1 final + 7.20 kg * v2 final
2.774 kg·m/s = 2.20 kg * v1 final + 7.20 kg * v2 final
Since the collision is elastic, kinetic energy is conserved. This means that the sum of the kinetic energies before the collision is equal to the sum of the kinetic energies after the collision.
Using the kinetic energy equation:
Initial kinetic energy = Final kinetic energy
(1/2) * m1 * (v1 initial)^2 + (1/2) * m2 * (v2 initial)^2 = (1/2) * m1 * (v1 final)^2 + (1/2) * m2 * (v2 final)^2
(1/2) * 2.20 kg * (1.27 m/s)^2 + (1/2) * 7.20 kg * (0 m/s)^2 = (1/2) * 2.20 kg * (v1 final)^2 + (1/2) * 7.20 kg * (v2 final)^2
1.737 kg·m^2/s^2 + 0 kg·m^2/s^2 = 1.10 kg·m/s + 7.20 kg·m/s
1.737 kg·m^2/s^2 = 8.30 kg·m/s
Now we have a system of two equations with two variables:
2.774 kg·m/s = 2.20 kg * v1 final + 7.20 kg * v2 final
1.737 kg·m^2/s^2 = 8.30 kg·m/s
We can solve this system of equations to find the values of v1 final and v2 final, the final velocities of the two balls.