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) 2.1 m/s (right) is the velocity of the blue ball after the collision.
Looks good
To check your answers, let's go through each question one by one:
1. The question provides the velocities of the two stones after the collision, but we need to determine the initial velocity of the second stone. To find the initial velocity, we can use the principle of conservation of momentum. The momentum before the collision should be equal to the momentum after the collision.
To find the initial velocity of the second stone, we need to break down the given velocities into their vector components. The given velocity of the first stone is 0.92 m/s [N71oW], and the given velocity of the second stone is 1.25 m/s [N44oE].
Using trigonometry, we can find the vertical and horizontal components of the velocities. The vertical component of the first stone's velocity is given by 0.92 sin(71°) and the horizontal component is given by 0.92 cos(71°). Similarly, the vertical component of the second stone's velocity is 1.25 sin(44°) and the horizontal component is 1.25 cos(44°).
Next, we can apply the conservation of momentum equation:
(mass of first stone × initial velocity of first stone) + (mass of second stone × initial velocity of second stone) = (mass of first stone × final velocity of first stone) + (mass of second stone × final velocity of second stone)
Since the masses of the stones are equal, we can simplify the equation:
(initial velocity of first stone) + (initial velocity of second stone) = (final velocity of first stone) + (final velocity of second stone)
We can substitute the given values and solve for the initial velocity of the second stone:
initial velocity of second stone = (final velocity of first stone) + (final velocity of second stone) - (initial velocity of first stone)
Substituting the given values:
initial velocity of second stone = 0.92 m/s - (1.25 cos(44°)) + (0.92 cos(71°))
Evaluating this expression will give you the answer.