# Collision and conservation of mechanical energy – probems and solutions

1. Two objects have the same mass

_{1}= m

_{2 }= 0.5 kg dropped from the same height as shown in the figure below. The radius of the circle is 1/5 m. The collision between both objects is perfectly elastic. Determine the velocity of each object after the collision. Acceleration due to gravity is 10 m/s

^{2}.

__Known :__

Mass of object (m) = m_{1 }= m_{2} = 0.5 kg

Initial height (h_{1}) = 1/5 m

Final height (h_{2}) = 0 *(**base of path**)*

Initial speed of object (v_{1}) = 0 *(**object initially at rest**)*

Final speed of object (v_{2}) = …. *(**speed of object at base of path** = **speed of object before collision**)*

Acceleration due to gravity (g) = 10 m/s

__Wanted :__ Speed of each object after collision

__Solution :__

**Speed of object before collision**

Speed of object before collision = speed of object when arrive at base of path = the final speed of object.

*Initial mechanical energy** = **final mechanical energy*

*The gravitational potential energy **+ **kinetic energy** = **the gravitational potential energy **+ **kinetic energy*

*m g h*_{1}* + 1/2 m v*_{1}^{2 }*= m g h*_{2}* + 1/2 m v*_{2}^{2 }

*m g h*_{1}* + 0** **= 0 + 1/2 m v*_{2}^{2 }

*m g h*_{1 }*= 1/2 m v*_{2}^{2 }

*g h*_{1} *= 1/2 v*_{2}^{2 }

*2 g h*_{1 }*= v*_{2}^{2 }

*2(10)(1/5) = v*_{2}^{2 }* *

*2(2) = v*_{2}^{2 }* *

*4 = v*_{2}^{2}

*v*_{2 }*= **√4*

*v*_{2} *= **2 m/s*

*Speed of each object before collision is **2 m//s.*

**The speed of object after collision**

If both objects have the same mass and move in opposite direction then when collide, both objects change its velocity. For example, if before collision object A moves at 2 m/s and object B moves at -4 m/s

2. Mass of object A is 2-kg and mass of object B is 3 kg dropped from a height as shown in the figure below. Both objects collide at point C. The collision is perfectly elastic. Acceleration due to gravity is 10 m/s^{2}. Determine the speed of object A and speed of object B after the collision.

__Known :__

Mass of object A (m_{1}) = 2 kg

Mass of object B (m_{2}) = 3 kg

Initial height (h_{1}) = 5 meters

Final height (h_{2}) = 0 *(**base of path**)*

Initial speed of the object (v_{1}) = 0 *(**initially object at rest**)*

Final speed of the object (v_{2}) = …. *(**speed at base of path** = **speed before collision**)*

Acceleration due to gravity (g) = 10 m/s

__Wanted :__ Speed of each object after collision

__Solution :__

**Speed of object before collision**

Speed of object before collision = speed of object at base of path

*The initial mechanical energy** = **the final mechanical energy*

*The gravitational potential energy** = **kinetic energy*

*m g h*_{1 }*= 1/2 m v*_{2}^{2 }

*g h*_{1 }*= 1/2 v*_{2}^{2 }

*2 g h*_{1} *= v*_{2}^{2 }

*2(10)(5) = v*_{2}^{2 }* *

*100 = v*_{2}^{2 }* *

*v*_{2 }*= **√100*

*v*_{2 }*= 10 **m/s*

*Speed of each object before collision **= 10 m/s. *

**Speed of object after collision**

Both objects have the different mass and moves in different direction so speed of each object after collision calculated using this equation.

The speed of each object just after the collision :