A 45 kg cart is moving across a frictionless floor at 1.7 m/s. A 60 kg boy, riding in the cart, jumps off the cart so that he hits the floor moving forwards in the initial direction of the cart's motion with a velocity of 2.8 m/s. What is the velocity of the cart after the boy jumped?
initial mass = 45 + 60 = 105 kg
initial momentum = 105 * 1.7 = final momentum
so
105*1.7 = 2.8*60 + 45 v
but how do you go from 105*1.7 = 2.8*60 + 45 v to the answer?? i'm so confused!
This is pretty basic algebra.
subtract 2.8*60 from each side. Then divide both sides by 45.
To solve this problem, we need to consider the principle of conservation of momentum. According to this principle, in the absence of external forces, the total momentum of a system remains constant.
The momentum of an object is given by the product of its mass and velocity. Therefore, to find the velocity of the cart after the boy jumped, we can use the equation:
(mass of cart * velocity of cart) + (mass of boy * velocity of boy) = (mass of cart * final velocity of cart) + (mass of boy * final velocity of boy)
Let's plug in the given values:
Mass of cart (m1) = 45 kg
Velocity of cart (v1) = 1.7 m/s
Mass of boy (m2) = 60 kg
Velocity of boy (v2) = 2.8 m/s
Final velocity of the cart (v1f) = ?
Final velocity of the boy (v2f) = 0 m/s (since the boy jumps off the cart and hits the floor)
Using the equation above, we have:
(45 kg * 1.7 m/s) + (60 kg * 2.8 m/s) = (45 kg * v1f) + (60 kg * 0 m/s)
Rearranging the equation to solve for v1f, we get:
(45 kg * 1.7 m/s) + (60 kg * 2.8 m/s) - (60 kg * 0 m/s) = (45 kg * v1f)
76.5 kg.m/s + 168 kg.m/s = 45 kg * v1f
244.5 kg.m/s = 45 kg * v1f
Now, divide both sides of the equation by 45 kg to solve for v1f:
v1f = 244.5 kg.m/s / 45 kg
v1f ≈ 5.43 m/s
Therefore, the velocity of the cart after the boy jumped is approximately 5.43 m/s.