A 3.0-kg cart is rolling along a frictionless, horizontal track towards a 1.2-kg cart that is held initially at rest. The carts are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is +3.9 m/s, and the second cart's velocity is -2.5 m/s. (a) What is the total momentum of the system of the two carts at this instant? (b) What was the velocity of the first cart when the second cart was still at rest?
(a) 3.0*3.9 - 1.2*2.5 = 8.7 kg*m/s
(b) Since TOTAL linear momentum must be conserved at all times before (and after)collision, when cart 2 is motionless, the speed of cart 1 must be 8.7/3.0 = 2.9 m/s.
Total momentum is conserved because the two carts interact magnetically with EACH OTHER rather than with an external magnet. The two forces on each other cancel.
A 198.7-kg and a 345.9-kg cart approach each other on a horizontal air track. They collide and stick together. After the collision their total kinetic energy is 1408.9 J. What is the speed after the collision? Only give the numerical value and don't include the unit
(a) Well, the total momentum of the system can be found by simply adding the individual momentum of each cart. The momentum of an object is calculated by multiplying its mass by its velocity. So, for the first cart with a mass of 3.0 kg and a velocity of +3.9 m/s, the momentum would be (3.0 kg)(3.9 m/s) = 11.7 kg·m/s. And for the second cart with a mass of 1.2 kg and a velocity of -2.5 m/s, the momentum would be (1.2 kg)(-2.5 m/s) = -3 kg·m/s. Now, to find the total momentum, simply add these two values together: 11.7 kg·m/s + (-3 kg·m/s) = 8.7 kg·m/s.
(b) To determine the velocity of the first cart when the second cart was at rest, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision. Since the second cart is initially at rest, its momentum is 0 kg·m/s. Therefore, the total momentum before the collision is equal to the momentum of the first cart, which we already found in part (a) to be 11.7 kg·m/s. So, using the same principle, we can set up the following equation: 11.7 kg·m/s = (3.0 kg + 1.2 kg) · v, where v is the velocity of the first cart when the second cart was at rest. Solving for v, we get v = 11.7 kg·m/s / 4.2 kg = 2.79 m/s. Therefore, the velocity of the first cart when the second cart was still at rest was approximately +2.79 m/s.
To find the solution, we need to first understand the concept of momentum, which is the product of an object's mass and velocity. The momentum of an object can be calculated using the equation:
Momentum (p) = Mass (m) × Velocity (v)
Now, let's solve the questions step by step:
(a) To find the total momentum of the system of the two carts at the given instant, we need to calculate the momentum of each cart separately and then sum them up.
Given:
Mass of the first cart (m1) = 3.0 kg
Velocity of the first cart (v1) = +3.9 m/s (positive because it is moving to the right)
Using the momentum equation, we can calculate the momentum of the first cart:
Momentum of the first cart (p1) = m1 × v1
p1 = 3.0 kg × 3.9 m/s = 11.7 kg·m/s
Given:
Mass of the second cart (m2) = 1.2 kg
Velocity of the second cart (v2) = -2.5 m/s (negative because it is moving to the left)
Using the momentum equation, we can calculate the momentum of the second cart:
Momentum of the second cart (p2) = m2 × v2
p2 = 1.2 kg × (-2.5 m/s) = -3.0 kg·m/s
To find the total momentum of the system, we simply sum up the individual momenta:
Total momentum = p1 + p2
Total momentum = 11.7 kg·m/s + (-3.0) kg·m/s = 8.7 kg·m/s
Therefore, the total momentum of the system of the two carts at the given instant is 8.7 kg·m/s.
(b) To determine the velocity of the first cart when the second cart was at rest, we can use the principle of conservation of momentum. According to this principle, the total momentum before and after a collision remains constant, assuming no external forces act on the system.
Since we are considering a frictionless track, we can apply the conservation of momentum in this scenario.
Momentum before collision = Momentum after collision
The first cart's momentum before the collision is given by:
p1_before = m1 × v1_before
Since we know p1_before = 11.7 kg·m/s (from part a) and m1 = 3.0 kg, we can solve for v1_before:
11.7 kg·m/s = 3.0 kg × v1_before
v1_before = 11.7 kg·m/s / 3.0 kg
v1_before ≈ 3.9 m/s
Therefore, the velocity of the first cart when the second cart was still at rest is approximately 3.9 m/s.