Question

A 282 kg bumper car moving 3.50 m/s collides with a 155 kg bumper car moving -1.38 m/s. Afterwards the 282 kg car moves at 1.10 m/s. Find the velocity of the 155 kg car afterwards?

Answer 1

Answer:

4.03 m/s

Explanation:

Initial momentum = final momentum

(282 kg) (3.50 m/s) + (155 kg) (-1.38 m/s) = (282 kg) (1.10 m/s) + (115 kg) v

v = 4.03 m/s

Answer 2

Answer:

$2.99\frac{m}{s} = v_{f_{2} }$

Explanation:

We know that from collisions (be it elastic or inelastic), as long as there are no external forces acting on the system, the momentum stays the same, before the collision and after the collision, this is called the Law of Conservation of Momentum.

The formula for momentum is:

M=mass * velocity

So from the law of conservation of momentum, we have:

Initial Momentum (before the collision) = Final momentum (after the collision)

If you check the simple drawing, you get the meaning of the sign in front of the velocity.

Now, using the conservation of momentum law's formula we have:

$m_{i_{1} } *v_{i_{1} } + m_{i_{2} }*v_{i_{2} } = m_{f_{1} }*v_{f_{1} }+m_{f_{2} }*v_{f_{2} }$

$282kg*3.50\frac{m}{s} + 155kg*(-1.38)\frac{m}{s} = 282kg*1.10\frac{m}{s} +155kg*v_{f_{2} }$

$987\frac{kg*m}{s} - 213.9*\frac{kg*m}{s} = 310.2\frac{kg*m}{s} +155kg*v_{f_{2} }$

$987\frac{kg*m}{s} - 213.9*\frac{kg*m}{s} = 310.2\frac{kg*m}{s} +155kg*v_{f_{2} }$

$462.9\frac{kg*m}{s} = 155kg*v_{f_{2} }$

$2.99\frac{m}{s} = v_{f_{2} }$ to the East