Respuesta :
4.) We are told that ball A is travelling from right to left, which we will refer to as a positive direction, making the initial velocity of ball A, +3 m/s. If ball B is travelling in the opposite direction to A, it will be travelling at -3 m/s. The final velocity of A is +2 m/s. Using the elastic collision equation, which uses the conservation of linear momentum, we can solve for the final velocity of B.
MaVai + MbVbi = MaVaf + MbVbf
Ma = 10 kg and Mb = 5 kg are the masses of balls A and B.
Vai = +3 m/s and Vbi = -3 m/s are the initial velocities.
Vaf = +2 m/s and Vbf = ? are the final velocities.
(10)(3) + (5)(-3) = (10)(2) + 5Vbf
30 - 15 = 20 + 5Vbf
15 = 20 + 5Vbf
-5 = 5 Vbf
Vbf = -1 m/s
The final velocity of ball B is -1 m/s.
5.) We are now told that Ma = Mb, but Vai = 2Vbi
We can use another formula to look at this mathematically.
Vaf = [(Ma - Mb)/(Ma + Mb)]Vai + [(2Mb/(Ma + Mb)]Vbi
Since Ma = Mb we can simplify this formula.
Vaf = [(0)/2Ma]Vai + [2Ma/2Ma]Vbi
Vaf = Vbi
Vbf = [(2Ma/(Ma + Mb)]Vai + [(Ma - Mb)/(Ma + Mb)]Vbi
Vbf = [2Mb/2Mb]Vai + [(0)/2Mb]Vbi
Vbf = Vai
Vaf = Vbi
Vbf = 2Vbi
If the initial velocity of A is twice the initial velocity of B, then the final velocity of A will be equal to the initial velocity of B.
If the initial velocity of A is twice the initial velocity of B, then the final velocity of B will be twice the initial velocity of B.
MaVai + MbVbi = MaVaf + MbVbf
Ma = 10 kg and Mb = 5 kg are the masses of balls A and B.
Vai = +3 m/s and Vbi = -3 m/s are the initial velocities.
Vaf = +2 m/s and Vbf = ? are the final velocities.
(10)(3) + (5)(-3) = (10)(2) + 5Vbf
30 - 15 = 20 + 5Vbf
15 = 20 + 5Vbf
-5 = 5 Vbf
Vbf = -1 m/s
The final velocity of ball B is -1 m/s.
5.) We are now told that Ma = Mb, but Vai = 2Vbi
We can use another formula to look at this mathematically.
Vaf = [(Ma - Mb)/(Ma + Mb)]Vai + [(2Mb/(Ma + Mb)]Vbi
Since Ma = Mb we can simplify this formula.
Vaf = [(0)/2Ma]Vai + [2Ma/2Ma]Vbi
Vaf = Vbi
Vbf = [(2Ma/(Ma + Mb)]Vai + [(Ma - Mb)/(Ma + Mb)]Vbi
Vbf = [2Mb/2Mb]Vai + [(0)/2Mb]Vbi
Vbf = Vai
Vaf = Vbi
Vbf = 2Vbi
If the initial velocity of A is twice the initial velocity of B, then the final velocity of A will be equal to the initial velocity of B.
If the initial velocity of A is twice the initial velocity of B, then the final velocity of B will be twice the initial velocity of B.
(4) It is given that :
Mass of ball A, [tex]m_A=10\ kg[/tex]
Mass of ball B, [tex]m_B=5\ kg[/tex]
Initial velocity of ball A, [tex]u_A=-3\ m/s[/tex]
Initial velocity of ball B, [tex]u_B=3\ m/s[/tex]
Final velocity of ball A, [tex]v_A=2m/s[/tex]
We have to find the final velocity of ball B after collision.
Since, it is a elastic collision. So the momentum remains conserved.
i.e.
[tex]m_Au_A+m_Bu_B=m_Av_A+m_Bv_B[/tex]
[tex]10\ kg\times (-3\ m/s)+5\ kg\times3\ m/s=10\ kg\times 2\ m/s+5\ kg\timesv_B[/tex]
[tex]-30+15-20=5v_B[/tex]
[tex]-35\ m/s=5v_B[/tex]
[tex]-7\ m/s=v_B[/tex]
So, after collision the ball B moves from right to left with a speed of 7 m/s.
(5) In this part, the mass of each ball is same and velocity of ball A is twice of that of ball B such that
[tex]u_A=2u_B[/tex]
[tex]m_A=m_B=m[/tex]
If the collision of two balls is elastic, then using the conservation of momentum as
[tex]3u_B=v_A+v_B[/tex]
This implies that the final velocities would be different.
If the collision is inelastic,
[tex]3v_b=2mv_f[/tex]
[tex]v_f=\dfrac{3v_b}{2}[/tex]
So, if there is inelastic collision the balls will move with a common velocities.