several students are riding in bumper cars at an amusement park. the combined mass of a car A and its occupants is 250 kg. the combined mass of car B and its occupants is 200 kg. Car A is 15 m away from car B and moving to the right at 2 m/s when the driver decides to bump into car B, which is at rest. car A accelerates at 1.5 m/s2 to a speed of 5 m/s and then continues at constant velocity until it strikes car B. Calucalte the total time for car A to travel the 15 m. After the collision car B moves to the right at a speed of 4.8 m/s. calculate the speed of car A after the collision. after the collision did car A move to the left, right, or is at rest? is this an elastic collision?
haha i had this same assignment at school like 2 weeks ago
(a) t = 2s, d = 7m, t = 1.6s, so overall t = 3.6s
(b)1.16 m/s to the right
(C)before: 3125, After:2449J, therefore not elastic
i did do my work! but i am tryin to confirm my answers. i appreciate the fact that you answered me. thank you.
To calculate the total time for car A to travel the 15 m, we can use the equation of motion:
Distance = Initial Velocity * Time + 0.5 * Acceleration * Time^2
Given:
Initial Velocity of car A, uA = 2 m/s
Acceleration of car A, aA = 1.5 m/s^2
Distance between car A and car B, d = 15 m
We will first calculate the time it takes for car A to accelerate from 2 m/s to 5 m/s:
5 m/s = 2 m/s + 1.5 m/s^2 * Time
Rearranging the equation:
Time = (5 m/s - 2 m/s) / 1.5 m/s^2
Time = 3 m/s / 1.5 m/s^2
Time = 2 seconds
Now we know that it takes 2 seconds for car A to accelerate to 5 m/s.
Next, we need to find the time it takes for car A to travel the remaining distance (15 m - 2 m * 2 seconds = 11 m) at a constant velocity of 5 m/s.
Distance = Velocity * Time
11 m = 5 m/s * Time
Solving for Time:
Time = 11 m / 5 m/s
Time = 2.2 seconds
Therefore, the total time for car A to travel the 15 m is 2 seconds (for acceleration) + 2.2 seconds (for constant velocity) = 4.2 seconds.
Now let's move on to the speed of car A after the collision and its direction:
Given:
Mass of car A and occupants, mA = 250 kg
Mass of car B and occupants, mB = 200 kg
Initial velocity of car B = 0 m/s
Final velocity of car B after the collision, vB = 4.8 m/s
To find the speed of car A after the collision, we can use the principle of conservation of momentum:
mAuA + mBuB = mAvA + mBvB
Since car B is at rest initially (uB = 0), the equation simplifies to:
mAuA = mAvA + mBvB
Rearranging the equation and substituting the known values:
mAuA - mAvA = mBvB
250 kg * 5 m/s - 250 kg * vA = 200 kg * 4.8 m/s
Solving for vA:
1250 kg.m/s - 250 kg * vA = 960 kg.m/s
Subtracting 1250 kg.m/s from both sides:
-250 kg * vA = -290 kg.m/s
Dividing both sides by -250 kg:
vA = (-290 kg.m/s) / (-250 kg)
vA = 1.16 m/s
The speed of car A after the collision is 1.16 m/s.
Lastly, let's determine the direction of car A after the collision and whether it was an elastic collision or not:
Since the speed of car A after the collision is positive and greater than zero (1.16 m/s), we can conclude that car A is moving to the right.
To determine if the collision is elastic, we need to check if kinetic energy was conserved. In an elastic collision, both momentum and kinetic energy are conserved.
Since only the final velocity of car B is given, we do not have enough information to determine if kinetic energy is conserved. Therefore, we cannot conclude if the collision is elastic or not.