Question

A railroad car of mass 2.50*10^4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with an initial speed of 2.00 m/s. (a) What is the speed of the four cars after the collision? (b) How much mechanical energy is lost in the collision?

Answer 1

Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

$mV_1+3mV_2=(m+3m)V_f=4mV_f$

$V_1+3V_2=4V_f$ and making $V_f$ the subject then

$V_f=\frac {V_1+3V_2}{4}$ and since $V_1$ is initial velocity of car, value given as 4 m/s, $V_2$ is the initial velocity of the three cars stuck together, value given as 2 m/s and $v_f$ is the final velocity which is unknown. By substitution

$V_f=\frac {4+(3\times2)}{4}=2.5 m/s$

(b)

Initial kinetic energy is given by

$\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ$

Final kinetic energy is given by

$\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ$

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

Answer 2

a) The speed of the four cars after the collision is 2.50 m/s

b) The loss in mechanical energy is $3.7\cdot 10^4 J$

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:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$  

where:  

$m_1 = 2.50\cdot 10^4 kg$ is the mass of the car

$u_1 = 4.00 m/s$ is the initial velocity of the car

$m_2 = 3m_1 = 3(2.50\cdot 10^4) = 7.50\cdot 10^4 kg$ is the mass of the other three cars, that move together at the same velocity (so we can consider them as a single object)

$u_2 =2.00 m/s$ is the initial velocity of the other three cars

$v$ is the final combined velocity of the four cars after coupling

Solving the equation for v, we find their final speed after the collision:

$v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(2.50\cdot 10^4)(4.00)+(7.50\cdot 10^4)(2.00)}{2.50\cdot 10^4+7.50\cdot 10^4}=2.50 m/s$

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:

$E_i = K_1 + K_2 = \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2u_2^2=\frac{1}{2}(2.50\cdot 10^4)(4.00)^2+\frac{1}{2}(7.50\cdot 10^4)(2.00)^2=3.50\cdot 10^5 J$

The total mechanical energy of the system after the collision is equal to the kinetic energy of the four cars coupled together, so:

$E_f = \frac{1}{2}(m_1 +m_2)v^2=\frac{1}{2}(2.50\cdot 10^4 + 7.50\cdot 10^4)(2.50)^2=3.13\cdot 10^5 J$

So, the loss of mechanical energy is

$-\Delta E=E_i -E_f = 3.50\cdot 10^5 - 3.13\cdot 10^5 = 3.7\cdot 10^4 J$

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