I'm preparing for a test. Can anyone check my answer for all these questions? These are my questions:
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.
1) 1.20 m/s (West 89.9 degrees North)
2) 1.6275 (kg.m)/s (East 49.6 degrees South) and the force exerted is 226N (East 49.6 degrees South)
3) 2.1m/s (right)
See previous post: Tue, 1-26-16, 1:52 PM.
To check your answers for these physics questions, I will explain the method to arrive at the correct answers.
1. Initial velocity of the second stone:
In this question, we have a glancing collision between two curling stones. We know the final velocities of both stones after the collision. To find the initial velocity of the second stone, we need to use the law of conservation of momentum.
Momentum before collision = Momentum after collision
The momentum of an object is given by the product of its mass and velocity. Let's denote the initial velocity of the second stone as V.
For the first stone:
Momentum before collision = mass of the first stone x initial velocity of the first stone
For the second stone:
Momentum before collision = mass of the second stone x initial velocity of the second stone
Since the masses of both stones are equal, we can equate the two expressions.
We have the following information:
Initial velocity of the first stone = 0 (as it is stationary)
Final velocity of the first stone = 0.92 m/s [N71oW]
Final velocity of the second stone = 1.25 m/s [N44oE]
Using vector addition, we can break down each velocity into its components along the north-south and east-west directions. The angle given indicates the direction from the north, measured clockwise.
Converting angles to compass bearings:
[N71oW] can be represented as approximately [West 19 degrees North]
[N44oE] can be represented as approximately [East 44 degrees North]
Let's calculate the components of these velocities:
For the first stone:
Velocity in the east-west direction = 0
Velocity in the north-south direction = 0.92 m/s
For the second stone:
Velocity in the east-west direction = 1.25 m/s x cos(44°)
Velocity in the north-south direction = 1.25 m/s x sin(44°)
Now, equating the momenta before the collision:
(mass of the first stone) x (initial velocity of the first stone) = (mass of the second stone) x (initial velocity of the second stone)
Plugging in the values:
(1) x (0) = (1) x (V)
Simplifying the equation, we find that the initial velocity of the second stone (V) is 0 m/s.
Therefore, the initial velocity of the second stone is 0 m/s.
To check your answer, make sure that you have correctly calculated the components of the given velocities and applied the law of conservation of momentum.
You can use this information to review and verify your answer for the first question.
Do you want me to explain the solution to the other questions as well?