A cart labeled A which is 5.00kg is rolling to the right at 15.0 m/s and impacts cart B that is completely still and weighs 2.00kg. The impulse of the collision on cart B was 20.0 Ns. What was the final velocity of the cart A and in which direction. Prove the conservation of momentum for this collision.
Since impulse is equal to the change in momentum of an object, we can say delta p = 20.0Ns on cart B and delta p = -20Ns on cart A. So, -20Ns = m(A)vf-m(B)vi. Plugging in values given gives us that -20Ns = 5kg(vf)-5kg(15m/s). Solving for vf gives us 11m/s and we can prove this is conserving the law of conservation of momentum by saying delta p(A) = -delta p(B). Plugging in values then shows that 20Ns = 20 Ns. The impulse or change of momentum in each cart was equal, proving this law.
x^2 + 7x = -12+y. Solve for this function's graph x-intercepts.
After moving this -12 to the other side of the equation, we can use a punnett square to solve for two multiples in the equation and find what x-values result in the product equaling zero. So, after using a square we find that x^2 + 7x = -12+y is equivalent to x^2 + 7x +12 = y which equals (x+3)(x+4)=0 since y would be 0 if we are looking for an x intercept. Then, by plugging in x=-3 and x=-4 we find that both result in y=0, being our x-intercepts.
What is the integral of between the functions y=x^2 and y=x from their intersection to the origin?
After graphing both functions, you will find their intersection to be at (1,1). Since we could integrate either in terms of x or y, I will choose in terms of x because the graph's shape makes this approach simpler. y=x is above y=x^2 so in order to find the area between the two functions, we will subtract the integral of each. This will give the integral of (x-x^2) dx. Since the edge of this integral is when x=1, we will integrate (x-x^2) and then plug x=1 into (x^2/2 -x^3/3) since plugging in 0 to this function and subtracting it with result in the same answer. The final result of the integral between these two functions is 1/6.