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)
1. To find the initial velocity of the second stone, we can use vector addition.
First, let's break down the given velocities into their x and y components.
For the first stone:
Velocity = 0.92 m/s [N71oW]
To find the x-component of the velocity:
Vx1 = V1 * cos(71)
Vx1 = 0.92 * cos(71)
Vx1 ≈ 0.35 m/s [E]
To find the y-component of the velocity:
Vy1 = V1 * sin(71)
Vy1 = 0.92 * sin(71)
Vy1 ≈ 0.88 m/s [N]
Similarly, for the second stone:
Velocity = 1.25 m/s [N44oE]
To find the x-component of the velocity:
Vx2 = V2 * cos(44)
Vx2 = 1.25 * cos(44)
Vx2 ≈ 0.89 m/s [E]
To find the y-component of the velocity:
Vy2 = V2 * sin(44)
Vy2 = 1.25 * sin(44)
Vy2 ≈ 0.92 m/s [N]
Now, let's add the x-components and y-components separately to find the resultant velocity:
For the x-component:
Vx = Vx1 + Vx2
Vx = 0.35 + 0.89
Vx ≈ 1.24 m/s [E]
For the y-component:
Vy = Vy1 + Vy2
Vy = 0.88 + 0.92
Vy ≈ 1.80 m/s [N]
Using the Pythagorean theorem:
Resultant velocity = √(Vx^2 + Vy^2)
Resultant velocity = √(1.24^2 + 1.80^2)
Resultant velocity ≈ 2.21 m/s
Therefore, the initial velocity of the second stone was approximately 2.21 m/s [N38oE].
2. To determine the impulse on the ball and the average force exerted during the collision, we can use the impulse-momentum principle.
Impulse = Change in momentum
First, let's calculate the initial momentum of the ball:
Momentum = mass * velocity
Initial momentum = 0.62 kg * 2.0 m/s [N]
Initial momentum = 1.24 kg·m/s [N]
Next, let's calculate the final momentum of the ball:
Final momentum = 0.62 kg * 1.7 m/s [E]
Final momentum = 1.054 kg·m/s [E]
Change in momentum = Final momentum - Initial momentum
Change in momentum = 1.054 kg·m/s [E] - 1.24 kg·m/s [N]
Change in momentum ≈ -0.186 kg·m/s [NE]
Impulse = Change in momentum
Impulse ≈ -0.186 kg·m/s [NE]
To find the average force exerted on the ball during the collision, we can use the equation:
Average force = Impulse / Duration of collision
Average force = -0.186 kg·m/s [NE] / 0.0072 s
Average force ≈ -25.83 N [NE]
Therefore, the impulse on the ball was approximately -0.186 kg·m/s [NE], and the average force exerted on the ball during the collision was approximately -25.83 N [NE].
3. In a head-on collision, the law of conservation of momentum applies. According to this law, the total momentum before and after the collision remains the same.
Let's denote the initial velocity of the red ball as Vr and the initial velocity of the blue ball as Vb.
Initial momentum = Final momentum
(mass of red ball * Vr) + (mass of blue ball * Vb) = (mass of red ball * final velocity of red ball) + (mass of blue ball * final velocity of blue ball)
Given:
Mass of red ball = Mass of blue ball
Initial velocity of red ball = 2.1 m/s [right]
Initial velocity of blue ball = -3.0 m/s [left]
Final velocity of red ball = -3.0 m/s [left]
Using the given values, we can rewrite the equation as:
(Mass * 2.1 m/s) + (Mass * -3.0 m/s) = (Mass * -3.0 m/s) + (Mass * Vb)
Mass and Mass cancel out from both sides:
2.1 m/s + (-3.0 m/s) = (-3.0 m/s) + Vb
Simplifying further:
-0.9 m/s = Vb
Therefore, the velocity of the blue ball after the collision was -0.9 m/s [right].
4. To determine the final velocity of the combined vehicles, we can again use the law of conservation of momentum.
Total momentum before the collision = Total momentum after the collision
Given:
Mass of car = 1800 kg
Mass of truck = 3200 kg
Initial velocity of car = 22 m/s [N]
Initial velocity of truck = 14 m/s [E]
Let's denote the final velocity of the combined vehicles as Vf.
Using the given values, the equation can be written as:
(mass of car * initial velocity of car) + (mass of truck * initial velocity of truck) = (mass of car + mass of truck) * Vf
Applying the equation:
(1800 kg * 22 m/s [N]) + (3200 kg * 14 m/s [E]) = (1800 kg + 3200 kg) * Vf
Simplifying further:
39600 kg·m/s [N] + 44800 kg·m/s [E] = 5000 kg * Vf
Adding up the momentum:
8400 kg·m/s = 5000 kg * Vf
Dividing by the total mass:
Vf = 8400 kg·m/s / 5000 kg
Vf ≈ 1.68 m/s
Therefore, the final velocity of the combined vehicles after the collision is approximately 1.68 m/s [right].
1. To solve this problem, we can use vector addition and subtraction.
First, let's analyze the initial velocity of the second stone. We know that the final velocity of the first stone is 0.92 m/s [N71oW], which means it is moving in the northwest direction.
Similarly, we know that the final velocity of the second stone is 1.25 m/s [N44oE], which means it is moving in the northeast direction.
To find the initial velocity of the second stone, we need to add the two final velocities in order to cancel out the direction components.
Using vector addition, we add the magnitudes of the final velocities.
Final velocity of the first stone = 0.92 m/s [N71oW]
Final velocity of the second stone = 1.25 m/s [N44oE]
Adding the magnitudes:
0.92 m/s + 1.25 m/s = 2.17 m/s
Next, we need to determine the direction of the resultant velocity. We can do this by finding the angle between the x-axis and the resultant vector.
To find the angle, we subtract the two angle values:
71o - 44o = 27o
Therefore, the initial velocity of the second stone is 2.17 m/s [27o].
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.