{"id":2324,"date":"2018-05-01T02:22:08","date_gmt":"2018-04-30T18:22:08","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2324"},"modified":"2023-08-08T12:41:16","modified_gmt":"2023-08-08T12:41:16","slug":"collision-and-conservation-of-mechanical-energy-probems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/collision-and-conservation-of-mechanical-energy-probems-and-solutions.htm","title":{"rendered":"Collision and conservation of mechanical energy \u2013 probems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Collision and conservation of mechanical energy \u2013 probems and solutions<\/span><\/p>\n

1. Two objects have the same mass<\/a>, m1<\/sub> = m2 <\/sub>= 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<\/a> between both objects is perfectly elastic. Determine the velocity<\/a> of each object after the collision. Acceleration due to gravity<\/a> is 10 m\/s2<\/sup>.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass of object (m) = m1 <\/sub>= m2<\/sub> = 0.5 kg\"Collision<\/span><\/p>\n

Initial height (h1<\/sub>) = 1\/5 m<\/span><\/p>\n

Final height (h2<\/sub>) = 0 (<\/i>base of path<\/i>)<\/i><\/span><\/p>\n

Initial speed of object (v1<\/sub>) = 0 (<\/i>object initially at rest<\/i>)<\/i><\/span><\/p>\n

Final speed of object (v2<\/sub>) = …. (<\/i>speed of object at base of path<\/i> = <\/i>speed of object before collision<\/i>)<\/i><\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s<\/span><\/p>\n

Wanted :<\/u> Speed of each object after collision<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Speed of object before collision<\/b><\/span><\/p>\n

Speed of object before collision = speed of object when arrive at base of path = the final speed of object.<\/span><\/p>\n

Initial mechanical energy<\/a><\/i> = <\/i>final mechanical energy<\/i><\/span><\/p>\n

The gravitational potential energy <\/a><\/i>+ <\/i>kinetic energy<\/i><\/a> = <\/i>the gravitational potential energy <\/i>+ <\/i>kinetic energy<\/i><\/span><\/p>\n

m g h<\/i>1<\/i><\/sub> + 1\/2 m v<\/i>1<\/i><\/sub>2 <\/i><\/sup>= m g h<\/i>2<\/i><\/sub> + 1\/2 m v<\/i>2<\/i><\/sub>2 <\/i><\/sup><\/span><\/p>\n

m g h<\/i>1<\/i><\/sub> + 0<\/i> <\/i>= 0 + 1\/2 m v<\/i>2<\/i><\/sub>2 <\/i><\/sup><\/span><\/p>\n

m g h<\/i>1 <\/i><\/sub>= 1\/2 m v<\/i>2<\/i><\/sub>2 <\/i><\/sup><\/span><\/p>\n

g h<\/i>1<\/sub> <\/i>= 1\/2 v<\/i>2<\/i><\/sub>2 <\/i><\/sup><\/span><\/p>\n

2 g h<\/i>1 <\/i><\/sub>= v<\/i>2<\/i><\/sub>2 <\/i><\/sup><\/span><\/p>\n

2(10)(1\/5) = v<\/i>2<\/i><\/sub>2 <\/i><\/sup> <\/i><\/span><\/p>\n

2(2) = v<\/i>2<\/i><\/sub>2 <\/i><\/sup> <\/i><\/span><\/p>\n

4 = v<\/i>2<\/i><\/sub>2<\/i><\/sup><\/span><\/p>\n

v<\/i>2 <\/i><\/sub>= <\/i>\u221a4<\/i><\/span><\/p>\n

v<\/i>2<\/sub> <\/i>= <\/i>2 m\/s<\/i><\/span><\/p>\n

Speed of each object before collision is <\/i>2 m\/\/s.<\/i><\/span><\/p>\n

The speed of object after collision<\/b><\/span><\/p>\n

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, after collision object A moves at 4 m\/s and object B moves -2 m\/s. Minus and plus sign indicates that both objects have the different direction.<\/span><\/p>\n

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\/s2<\/sup>. Determine the speed of object A and speed of object B after the collision.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass of object A (m1<\/sub>) = 2 kg\"Collision<\/span><\/p>\n

Mass of object B (m2<\/sub>) = 3 kg<\/span><\/p>\n

Initial height (h1<\/sub>) = 5 meters<\/span><\/p>\n

Final height (h2<\/sub>) = 0 (<\/i>base of path<\/i>)<\/i><\/span><\/p>\n

Initial speed of the object (v1<\/sub>) = 0 (<\/i>initially object at rest<\/i>)<\/i><\/span><\/p>\n

Final speed of the object (v2<\/sub>) = …. (<\/i>speed at base of path<\/i> = <\/i>speed before collision<\/i>)<\/i><\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s<\/span><\/p>\n

Wanted :<\/u> Speed of each object after collision<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Speed of object before collision<\/b><\/span><\/p>\n

Speed of object before collision = speed of object at base of path<\/span><\/p>\n

The initial mechanical energy<\/i> = <\/i>the final mechanical energy<\/i><\/span><\/p>\n

The gravitational potential energy<\/i> = <\/i>kinetic energy<\/i><\/span><\/p>\n

m g h<\/i>1 <\/i><\/sub>= 1\/2 m v<\/i>2<\/i><\/sub>2 <\/i><\/sup><\/span><\/p>\n

g h<\/i>1 <\/i><\/sub>= 1\/2 v<\/i>2<\/i><\/sub>2 <\/i><\/sup><\/span><\/p>\n

2 g h<\/i>1<\/sub> <\/i>= v<\/i>2<\/i><\/sub>2 <\/i><\/sup><\/span><\/p>\n

2(10)(5) = v<\/i>2<\/i><\/sub>2 <\/i><\/sup> <\/i><\/span><\/p>\n

100 = v<\/i>2<\/i><\/sub>2 <\/i><\/sup> <\/i><\/span><\/p>\n

v<\/i>2 <\/i><\/sub>= <\/i>\u221a100<\/i><\/span><\/p>\n

v<\/i>2 <\/i><\/sub>= 10 <\/i>m\/s<\/i><\/span><\/p>\n

Speed of each object before collision <\/i>= 10 m\/s. <\/i><\/span><\/p>\n

Speed of object after collision<\/b><\/span><\/p>\n

Both objects have the different mass and moves in different direction so speed of each object after collision calculated using this equation.<\/span><\/p>\n

\"Collision<\/span><\/p>\n

The speed of each object just after the collision :<\/span><\/p>\n

\"Collision<\/span><\/p>\n

    \n
  1. What is the difference between an elastic and an inelastic collision?<\/strong><\/span>\n