Answer:
[tex]v = 5.833\,\frac{m}{s}[/tex]
Explanation:
The collision is inelastic and can be described by the Principle of Momentum Conservation:
[tex](700\,kg)\cdot (10\,\frac{m}{s} ) + (500\,kg)\cdot (0\,\frac{m}{s} ) = (1200\,kg)\cdot v[/tex]
The speed after the collision is:
[tex]v = 5.833\,\frac{m}{s}[/tex]
Answer:
5.83 m/s
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision.
mu+m'u' = V(m+m')..................... Equation 1
Where m = mass of the first car, m' = mass of the second car, u = initial velocity of the first car, u' = initial velocity of the second car, V = common velocity of both cars after collision
make V the subject of the equation,
V = (mu+m'u')/(m+m')................. Equation 2
Given: m = 700 kg, m' = 500 kg, u = 10 m/s, u' = 0 m/s ( at rest)
Substitute into equation 2
V = (700×10+500×0)/(700+500)
V = 7000/1200
V = 5.83 m/s
