1. Object with mass of 75kg slides down from rest. what is the height of the ini
ID: 2020077 • Letter: 1
Question
1. Object with mass of 75kg slides down from rest. what is the height of the initial position of the object if the speed of 5 m/s at the bottom?(ignore friction)m=75kg
Vo=0
V= 5 m/s
2. Railroad car of mass 20000kg moving with a speed of 4 m/s collides with another car of the same mass moving in the same direction with speed of 1.5 m/s after collision they are coupled together. what is the speed of the coupled cars?
M1=20000
V1=4 m/s
M2=20000
V2= 1.5 m/s
3. An 80g bullet is fired into a 3 kg ballistic pendulum and is embedded inside it as a result of collision. what is the speed of the system right after the collision? At what height will the pendulum with a bullet inside it rise? The initial velocity of the bullet was 7 m/s.
V=
Vo= 7m/s
M1=80 g
M2= 2kg
Explanation / Answer
1) Given that, Mass of the object m = 75 kg Initial velocity of the object u = 0 m/s Final velocity of the object v = 5 m/s Let the initial height = h When the object starts from a height h and reached the bottom all its potential energy converts into kinetic energy. So, mgh = (1 / 2)mv2 gh = (1 / 2)v2 h = 0.5v2 / g h = 1.275 m 2) Mass of the first car m1 = 20000kg Mass of the second car m2 = 20000kg Initial velocity of the first car v1 = 4 m/s Initial velocity of the second car v2 = 1.5 m/s Let the final velocity of the combined system after collision = V According to conservation of momentum m1v1 +m2v2 = (m1 + m2 )V V = (m1v1 +m2v2 )/ (m1 + m2 ) V = 2.75 m/s 3) Mass of the bullet m1 = 80 g = 0.08 kg Mass of the pendulam m2 = 3 kg Initial velocity of the bullet u1 = 7 m/s Initial velocity of the pendulam u2 = 0 m/s According to conservation of momentum m1u1 +m2u2 = ( m1 + m2 )V V = (m1u1 +m2u2)/ (m1 + m2 ) V = 0.1818 m/s At certain height h , (1/2) (m1 + m2 )V2 = ( m1 + m2 )gh h = V2 / 2g h = 0.00168 m gh = (1 / 2)v2 h = 0.5v2 / g h = 1.275 m 2) Mass of the first car m1 = 20000kg Mass of the second car m2 = 20000kg Initial velocity of the first car v1 = 4 m/s Initial velocity of the second car v2 = 1.5 m/s Let the final velocity of the combined system after collision = V According to conservation of momentum m1v1 +m2v2 = (m1 + m2 )V V = (m1v1 +m2v2 )/ (m1 + m2 ) V = 2.75 m/s 3) Mass of the bullet m1 = 80 g = 0.08 kg Mass of the pendulam m2 = 3 kg Initial velocity of the bullet u1 = 7 m/s Initial velocity of the pendulam u2 = 0 m/s According to conservation of momentum m1u1 +m2u2 = ( m1 + m2 )V V = (m1u1 +m2u2)/ (m1 + m2 ) V = 0.1818 m/s At certain height h , (1/2) (m1 + m2 )V2 = ( m1 + m2 )gh h = V2 / 2g h = 0.00168 m 3) Mass of the bullet m1 = 80 g = 0.08 kg Mass of the pendulam m2 = 3 kg Initial velocity of the bullet u1 = 7 m/s Initial velocity of the pendulam u2 = 0 m/s According to conservation of momentum m1u1 +m2u2 = ( m1 + m2 )V V = (m1u1 +m2u2)/ (m1 + m2 ) V = 0.1818 m/s At certain height h , (1/2) (m1 + m2 )V2 = ( m1 + m2 )gh h = V2 / 2g h = 0.00168 mRelated Questions
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