Answer :
Answer:
(a) 2.5 m/s
(b) 37.5 KJ
Explanation:
(a)
From the law of conservation of momentum, Initial momentum=Final momentum
[tex]mV_1+3mV_2=(m+3m)V_f=4mV_f[/tex]
[tex]V_1+3V_2=4V_f[/tex] and making [tex]V_f[/tex] the subject then
[tex]V_f=\frac {V_1+3V_2}{4}[/tex] and since [tex]V_1[/tex] is initial velocity of car, value given as 4 m/s, [tex]V_2[/tex] is the initial velocity of the three cars stuck together, value given as 2 m/s and [tex]v_f[/tex] is the final velocity which is unknown. By substitution
[tex]V_f=\frac {4+(3\times2)}{4}=2.5 m/s[/tex]
(b)
Initial kinetic energy is given by
[tex]\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ[/tex]
Final kinetic energy is given by
[tex]\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ[/tex]
The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence
Energy lost=350-312.5=37.5 KJ
a) The speed of the four cars after the collision is 2.50 m/s
b) The loss in mechanical energy is [tex]3.7\cdot 10^4 J[/tex]
Explanation:
a)
We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the system must be conserved before and after the collision.
Mathematically, for this problem we can write:
[tex]p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v[/tex]
where:
[tex]m_1 = 2.50\cdot 10^4 kg[/tex] is the mass of the car
[tex]u_1 = 4.00 m/s[/tex] is the initial velocity of the car
[tex]m_2 = 3m_1 = 3(2.50\cdot 10^4) = 7.50\cdot 10^4 kg[/tex] is the mass of the other three cars, that move together at the same velocity (so we can consider them as a single object)
[tex]u_2 =2.00 m/s[/tex] is the initial velocity of the other three cars
[tex]v[/tex] is the final combined velocity of the four cars after coupling
Solving the equation for v, we find their final speed after the collision:
[tex]v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(2.50\cdot 10^4)(4.00)+(7.50\cdot 10^4)(2.00)}{2.50\cdot 10^4+7.50\cdot 10^4}=2.50 m/s[/tex]
b)
The total mechanical energy of the system before the collision is equal to the kinetic energy of the first car + the kinetic energy of the other three cars, so:
[tex]E_i = K_1 + K_2 = \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2u_2^2=\frac{1}{2}(2.50\cdot 10^4)(4.00)^2+\frac{1}{2}(7.50\cdot 10^4)(2.00)^2=3.50\cdot 10^5 J[/tex]
The total mechanical energy of the system after the collision is equal to the kinetic energy of the four cars coupled together, so:
[tex]E_f = \frac{1}{2}(m_1 +m_2)v^2=\frac{1}{2}(2.50\cdot 10^4 + 7.50\cdot 10^4)(2.50)^2=3.13\cdot 10^5 J[/tex]
So, the loss of mechanical energy is
[tex]-\Delta E=E_i -E_f = 3.50\cdot 10^5 - 3.13\cdot 10^5 = 3.7\cdot 10^4 J[/tex]
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