Respuesta :
Answer:
A) [tex]v_b=0.8\ m.s^{-1}[/tex] is the final velocity of the cart B after collision.
B) [tex]KE_A=0.176\ J[/tex]
C) [tex]KE_B=0\ J[/tex]
D) [tex]ke_a=0\ J[/tex]
E) [tex]ke_B=0.176\ J[/tex]
F) Yes, here the kinetic energy is conserved because the mass of both the bodies involved in the collision is same.
G) Yes, momentum is always conserved for an elastic collision.
Explanation:
Given:
- mass of car A, [tex]m_a=0.55\ kg[/tex]
- mass of car B, [tex]m_b=0.55\ kg[/tex]
- initial velocity of car A, [tex]u_a=0.8\ m.s^{-1}[/tex]
- final velocity of the car A, [tex]v_a=0\ m.s^{-1}[/tex]
A)
As given in the question that the cars undergo an elastic collision:
According to the conservation of momentum:
[tex]m_a.u_a+m_b.u_b=m_a.v_a+m_b.v_b[/tex]
[tex]0.55\times 0.8+0.55\times 0=0.55\times 0+0.55\times v_b[/tex]
[tex]v_b=0.8\ m.s^{-1}[/tex] is the final velocity of the cart B after collision.
B)
Initial kinetic energy of cart A:
[tex]KE_A=\frac{1}{2} m_a.u_a^2[/tex]
[tex]KE_A=0.5\times 0.55\times 0.8^2[/tex]
[tex]KE_A=0.176\ J[/tex]
C)
Initial kinetic energy of cart A:
[tex]KE_B=\frac{1}{2} \times m_b.u_b^2[/tex]
[tex]KE_B=0.5\times 0.55\times 0^2[/tex]
[tex]KE_B=0\ J[/tex]
D)
The final kinetic energy of cart A:
[tex]ke_A=\frac{1}{2} m_a.v_a^2[/tex]
[tex]ke_a=0.5\times 0.55\times 0^2[/tex]
[tex]ke_a=0\ J[/tex]
E)
The final kinetic energy of cart B:
[tex]ke_B=\frac{1}{2} m_b.v_b^2[/tex]
[tex]ke_B=0.5\times 0.55\times 0.8^2[/tex]
[tex]ke_B=0.176\ J[/tex]
F)
Yes, here the kinetic energy is conserved because the mass of both the bodies involved in the collision is same.
G)
Yes, momentum is always conserved for an elastic collision.
(a) The velocity of cart B is [tex]0.8m/s[/tex]
(b) Initial kinetic energy of cart A is [tex]0.176J[/tex]
(c) Initial kinetic energy of cart B is 0
(d) The final kinetic energy of cart A is 0
(e) The final kinetic energy of cart B is [tex]0.176J[/tex]
(f) The kinetic energy is conserved
(g) The momentum is conserved
Elastic collision:
(a) The momentum of the system must be conserved, which means momentum before the collision must be equal to momentum after the collision.
Momentum before collision = [tex]m_Au_A+m_bu_B=0.55\times0.8+0.55\times0=0.44kgm/s[/tex]
Momentum after collision:
[tex]m_A+m_A+m_Bv_B=0.55\times0+0.55v_B=0.44\\\\v_B=0.8m/s[/tex]
(b) Initial kinetic energy of cart A:
[tex]K_{iA}=\frac{1}{2}m_Au_A^2=0.5\times0.55\times0.64=0.176J[/tex]
(c) Initial kinetic energy of cart B:
[tex]K_{iB}=\frac{1}{2}m_Bu_B^2=0.5\times0.55\times0=0J[/tex]
(d) The final kinetic energy of cart A:
[tex]K_{fA}=\frac{1}{2}m_Av_A^2=0.5\times0.55\times0=0J[/tex]
(e) Initial kinetic energy of cart B:
[tex]K_{iB}=\frac{1}{2}m_B_v_B^2=0.5\times0.55\times0.64=0.176J[/tex]
(f) yes the kinetic energy of the system is conserved for elastic collision
(g) The momentum is conserved in any case including the elastic collision
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