1. A stationary curling stone is struck in a glancing collision by a second curling stone of equal
mass. If the first stone moves away at a velocity of 0.92 m/s [N71oW] and the second stone
moves away at a velocity of 1.25 m/s [N44oE], what was the initial velocity of the second stone?
(5 marks)
2. A billiard ball (0.62 kg) with a velocity of 2.0 m/s [N] hits another ball and has a velocity of 1.7
m/s [E] after the collision. Determine the impulse on the ball and the average force exerted on it
during the collision if the duration of the collision was 0.0072 s. (5 marks)
3. Two billiard balls of equal mass undergo a head on collision. The red ball is travelling at 2.1
m/s [right] and hits the blue ball travelling at 3.0 m/s [left]. If the speed of the red ball after the
collision is 3.0 m/s [left], determine the velocity of the blue ball after the collision. (5 marks)
4. A car with a mass of 1800 kg is initially travelling with a velocity of 22 m/s [N] when it collides
with a truck with a mass of 3200 kg traveling with a velocity of 14 m/s [E]. If the two vehicles
become attached during the collision, determine their final velocity. (5 marks)
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Add your way on how to do this question. Am I correct?
1. Momentum before = after
In x net is 1.25cos44 - .92 cos71 = .6 in positive x
In y net is .92sin71 + 1.25sin44 = 1.74 positive y. Use pythagorean and tan-1 to find magnitude and direction
2.Ft = delta mv = .62(2-1.7) Divide t to get F.
3. Make left positive. Before p =m(3-2.1). After p = m(3-v). Is is obviously 2.1 right.
4. Use pythgorean to find vector sum momentum, then divide by total mass for velocity.
2 No. V is a vector, as is F. Do it as a vector equation, not as a scalar.
others look right. on 4, your method works, but I wouldn't make a practice, as this only works when the equation is perfectly inelastic.
bob, can you show me your perspective on how to approach these questions pls? Each question.
I would write momentum equations as vectors. 1 as an example. find the N, E components of Each vector, then write a momentum equation in the N direction, then write one in the E direction, and finally, an conservation of energy equation. You will have three equations, I only see two unknowns, the velocity and angle theta (which is the same as two velocity components). So three equations, two unknowns, some algebra involved.
For question one you don't need energy. Both stones are defined after with magnitude and direction. All of that momentum had to come from the second stone as the first was at rest. It's a simple addition of two momentum vectors.
Question 2 is indeed a vector equation, but it is all in one dimension. The impulse is just the change in momentum of the ball.
Question 4 explicitly states it is inelastic
--The REAL Chanz
1. Yes, you are correct. To find the initial velocity of the second stone, you can use the principle of conservation of momentum, which states that the total momentum before a collision is equal to the total momentum after the collision.
In this case, the momentum before the collision is given by the equation:
Momentum before = Momentum after
(mass of first stone * velocity of first stone) + (mass of second stone * velocity of second stone) = (mass of first stone * final velocity of first stone) + (mass of second stone * final velocity of second stone)
In the x-direction, you can calculate the net initial momentum:
In x net = (mass of second stone * initial velocity of second stone * cos(angle of initial velocity of second stone)) - (mass of first stone * initial velocity of first stone * cos(angle of initial velocity of first stone))
In the y-direction, you can calculate the net initial momentum:
In y net = (mass of first stone * initial velocity of first stone * sin(angle of initial velocity of first stone)) + (mass of second stone * initial velocity of second stone * sin(angle of initial velocity of second stone))
Using these equations, you can calculate the magnitude and direction of the initial velocity of the second stone by using the Pythagorean theorem and the arctangent function, respectively.
2. To find the impulse on the ball and the average force exerted on it during the collision, you can use the equation:
Impulse = change in momentum = mass of the ball * change in velocity
In this case, the change in velocity is given by the equation:
Change in velocity = final velocity - initial velocity
The impulse can then be calculated by multiplying the mass of the ball by the change in velocity.
Average force can be found by dividing the impulse by the duration of the collision.
3. Yes, you are correct. To find the velocity of the blue ball after the collision, you can apply the principle of conservation of momentum. The momentum before the collision is given by the equation:
Momentum before = Momentum after
In this case, the momentum before the collision is equal to the sum of the momentum of the red ball and the blue ball:
Momentum before = (mass of red ball * velocity of red ball) + (mass of blue ball * velocity of blue ball)
The momentum after the collision is given by the equation:
Momentum after = (mass of red ball * final velocity of red ball) + (mass of blue ball * final velocity of blue ball)
You can then solve for the final velocity of the blue ball by rearranging the equation and substituting the given values.
4. Yes, you are correct. To find the final velocity of the combined car and truck, you can use the principle of conservation of momentum. The momentum before the collision is given by the equation:
Momentum before = Momentum after
In this case, the momentum before the collision is equal to the sum of the momentum of the car and the truck:
Momentum before = (mass of car * velocity of car) + (mass of truck * velocity of truck)
The momentum after the collision is given by the equation:
Momentum after = (mass of car + mass of truck) * final velocity of combined vehicles
You can then solve for the final velocity of the combined vehicles by rearranging the equation and substituting the given values.