Respuesta :
Answer: velocity of big cart after collision is 0.00205m/s
Explanation: this question is based on conservation of linear momentum during collision.
This law states that during collision, the vector sum of momentum before collision equals the vector sum of collision after momentum.
Note p = m* v
Where p = linear momentum, m = mass of object and v = velocity.
From the question, the small cart has a mass of 100g=0.1kg with a velocity of 1.2m/s thus it momentum before collision is 0.1 * 1.2 = 0.12kgm/s
The big cart has a mass of 100kg but at rest meaning it velocity is zero (0), thus it momentum before collision is 100 * 0 = 0kgm/s
Thus the vector sum of momentum before collision is the momentum of the small and big cart which is
0.12 + 0 = 0.12kgm/s
Thus momentum before collision is 0.12kgm/s
After collision, the small cart moves but in the opposite direction (this is because it cannot push the big cart but rather it will hit it and bounce back in opposite direction) with a velocity of 0.850m/s, thus it momentum after collision is
- 0. 1 * 0.850 = - 0.0085kgm/s ( the negative sign is to show a change in direction since momentum is a vector quantity which is sensitive to directions)
The big cart moves with a velocity v after collision thus making it have a momentum of
100 * v = 100v
Thus the vector sum of momentum after collision is
100v - 0.0085
By using the law of conservation of momentum, momentum before collision equals that after collision, hence
0.12 = 100v - 0.085
By collecting like terms
0.12 + 0.085 = 100v
0.205 = 100v
v = 0.205/100
v= 0.00205kgm/s
The speed of the large cart after collision is 0.205 m/s.
To calculate the speed of the large cart after collision, we use the law of conservation of momentum
Law of conservation of momentum
States that for a system of colliding body, the total momentum before collision is equal to the total momentum after collision, provided there is not net external force acting on the body.
- Total momentum befor collision = Total mometum after collision
- mu+m'u' = mv+m'v'.......... Equation 1
Where:
- m = mass of the small cart
- u = initial velocity of the small cart
- m' = mass of the big cart
- u' = initial velocity of the big cart
- v = final velocity of the small cart
- v' = final velocity of the big cart
Fro the question,
Given:
- m = 100 g = 0.1 kg
- u = 1.2 m/s
- m' = 1 kg
- u' = 0 m/s (at rest)
- v = -0.85 m/sm (recoils)
Substitute these values into equation 1 and solve for v'
- (0.1×1.2)+(1×0) = (0.1×-0.850)+(1×v')
- 0.12 = -0.085+v'
- v' = 0.12+0.085
- v' = 0.205 m/s
Hence, The speed of the large cart after collision is 0.205 m/s.
Learn more law of conservation of momentum here: https://brainly.com/question/7538238
