Question

During hand-pumped rail car races, a speed of 27.9 km/h has been achieved by teams of four people. A car that has a mass equal to 379 kg is moving at that speed toward a river when Carlos, the chief pumper, notices that the bridge ahead is out. All four people (each with a mass of 75.0 kg) simultaneously jump backward off the car with a velocity that has a horizontal component of 4.00 m/s relative to the car. The car proceeds off the bank and falls into the water a horizontal distance of 22.1 m from the bank.

a) How long is the time of fall of the rail car?

b) What is the horizontal component of the velocity of the pumpers when they hit the ground?

Answer 1

Speed of the hand-pumped rail car is, u_ci = 27.9 km/h = (27000/3600) m/s = 7.5 m/s

Mass of the rail car is, m_c = 379 kg

Mass of each pumper is, m_p = 75 kg

Thus, total mass of the pumpers is, m_pt = 300 kg

Horizontal component of the velocity with which the pumpers jump backward relative to the car is, v_ph = -4 m/s [negative sign accounts for the direction relative to the direction of motion of the car]

Horizontal distance from the bank to the place where the car falls into water is, d = 22.1 m

(a)

From the conservation of momentum,

(m_c + m_pt) u_ci = m_c v_cf + m_pt v_pf

But,

v_pf - v_cf = v_ph

Thus,

v_pf= v_cf - 4

So,

(m_c + m_pt) u_ci = m_c v_cf + m_pt (v_cf - 4)

(m_c + m_pt) u_ci = (m_c + m_pt) v_cf - 4 * m_pt

Thus,

v_cf = [ (m_c + m_pt) u_ci + 4 * m_pt ] / (m_c + m_pt)

On substituting the values, we get,

v_cf = [ (379 + 300) * 7.5 + 4 * 300 ] / (379 + 300)

v_cf = 9.26730486 m/s = 9.267 m/s

Thus,

Time of fall of the rail car is,

t = d / v_cf

t = 22.1 / 9.26730486

t = 2.384727851 s = 2.385 s

(b)

Horizontal component of the velocity of the pumpers when they hit the ground is,

v_pf= v_cf - 4

v_pf= 9.26730486 - 4

v_pf= 5.26730486 m/s = 5.267 m/s