Question

A 212 kg bumper car is traveling 8.00 m/s when it rear ends a second 196 kg bumper car traveling at 6.75 m/s. If the cars get stuck together, how fast are they traveling after the collision?

Answer 1

Answer:

7.40m/s

Explanation:

The masses of the cars with their respective initial velocities are;

$m_1 = 212kg$

$u_1 = 8.00 m{s}^{ - 1}$

$m_2 = 196kg$

$u_1 =6.75m {s}^{ - 1}$

Since the two car stuck after collision, they all move with a common velocity in the same direction.

Let the velocity at which the two cars will be moving after collision be V.

From the conservation of momentum, in a closed system, momentum before collision is equal to momentum after collision.

Mathematically,

$m_1u_1 +m_2u_2 = (m_1 + m_2)v$

By substitution, we obtain;

$(212 \times 8) +(196 \times 6.75) = (212 + 196)v$

$\implies1696+1323 =408 \times v$

$\implies3019 =408 \times v$

Dividing through by 408, we obtain

$\implies\frac{3019}{408} = \frac{408 \times v }{408}$

$\implies v =7.3995$

Therefore,v=7.40m/s