Question

A 1,550 kg car moving south at 10.0 m/s collides with a 2,550 kg car moving north. The cars stick together and move as a unit after the collision at a velocity of 5.22 m/s to the north. Fins the velocityof the 2,550 kg car before the collision.

Answer 1

Answer:

14.47 m/s

Explanation:

The momentum must be preserved before and after the collision:

The total momentum before the collision

$m_1v_1 + m_2v_2$

where $m_1 = 1550 kg, m_2 = 2550 kg$ are the masses of the car moving south and north, respectively, before the collision. $v_1 = -10m/s$ is the velocity of the car moving South. We take the velocity to the North as the positive direction

The total momentum after the collision

$(m_1 + m_2)V$

where V = 5.22m/s is the velocity of both cars after the collision

We can equalize the 2 equations and plug in the numbers:

$m_1v_1 + m_2v_2 = (m_1 + m_2)V$

$1550(-10) + 2550*v_2 = (1550 + 2550)5.22$

$-15500 + 2550v_2 = 21402$

$2550v_2 = 36902$

$v_2 = 36902/2550 = 14.47 m/s$

Answer 2

The velocity of the 2,550 kg car before the collision is 14.47 m /s

The collision is a perfectly inelastic collision. In this case the maximum

amount of kinetic energy of the colliding object is lost.

Using the conservation of momentum and energy,

m₁u₁ + m₂u₂ = (m₁ + m₂)v

m₁ = 1550 kg

u₁ = - 10.0 m/s (travelling in the south direction)

m₂ = 2550 kg

u₂  = ?

v = 5.22 m/s

1550 × -10 + 2550 × u₂ = (1550 + 2550)5.22

-15500 + 2550u₂ = 4100 × 5.22

2550u₂ = 21402 + 15500

2550u₂ = 36902

u₂ = 36902  / 2550

u₂ = 14.47137

u₂ = 14.47  m /s

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