Two basketball players collide head on. play a weighs 80 kg and is travelling 2.5m/s to the right while player b weighs 68 kg and is travelling 1.2 m/s to the left. after collision player a is travelling at 1.0 m/s to the right.
a. what is the change in momentum of player a?
b. if the collision lasted 0.1s what is the average force play b must have exerted on player a during the collision?
c. what is the average force that player a must have exerted on player b during the collision?
d. what is the change in momentum of player b?
e. what is the final velocity of player b?
a. To find the change in momentum of player A, we can use the equation:
Δp = m * Δv, where Δp represents the change in momentum, m represents the mass, and Δv represents the change in velocity.
The initial velocity of player A is 2.5 m/s to the right, and the final velocity is 1.0 m/s to the right. Therefore, the change in velocity (Δv) is given by:
Δv = final velocity - initial velocity
Δv = 1.0 m/s - 2.5 m/s
Δv = -1.5 m/s (since it is in the opposite direction)
Now, we can calculate the change in momentum of player A:
Δp = m * Δv
Δp = 80 kg * (-1.5 m/s)
Δp = -120 kg·m/s
So, the change in momentum of player A is -120 kg·m/s.
b. To find the average force that player B must have exerted on player A during the collision, we can use the equation:
F = Δp / Δt, where F represents the force, Δp represents the change in momentum, and Δt represents the time duration of the collision.
The change in momentum of player A is -120 kg·m/s, and the time duration of the collision is 0.1 s. Therefore, we can calculate the average force:
F = Δp / Δt
F = -120 kg·m/s / 0.1 s
F = -1200 N
So, the average force that player B must have exerted on player A during the collision is -1200 N (since it is in the opposite direction).
c. To find the average force that player A must have exerted on player B during the collision, we can use Newton's third law of motion, which states that the forces exerted by two objects on each other are equal in magnitude but opposite in direction.
Therefore, the average force that player A exerted on player B during the collision is also 1200 N (in the opposite direction).
d. To find the change in momentum of player B, we can use the equation:
Δp = m * Δv, where Δp represents the change in momentum, m represents the mass, and Δv represents the change in velocity.
The initial velocity of player B is 1.2 m/s to the left, and the final velocity is the same. Therefore, the change in velocity (Δv) is given by:
Δv = final velocity - initial velocity
Δv = 1.2 m/s - 1.2 m/s
Δv = 0 m/s
The change in momentum of player B is then:
Δp = m * Δv
Δp = 68 kg * 0 m/s
Δp = 0 kg·m/s
So, the change in momentum of player B is 0 kg·m/s.
e. The final velocity of player B remains the same as its initial velocity, which is 1.2 m/s to the left.
a. The change in momentum of player A can be calculated by using the formula:
Change in momentum = mass of player A * (final velocity of player A - initial velocity of player A)
Given:
Mass of player A (mA) = 80 kg
Initial velocity of player A (uA) = 2.5 m/s
Final velocity of player A (vA) = 1.0 m/s
Change in momentum = 80 kg * (1.0 m/s - 2.5 m/s)
Change in momentum = 80 kg * (-1.5 m/s)
Change in momentum = -120 kg·m/s
Therefore, the change in momentum of player A is -120 kg·m/s.
b. To calculate the average force player B must have exerted on player A during the collision, we can use Newton's second law of motion:
Average force = Change in momentum / Time
Given:
Change in momentum = -120 kg·m/s
Time (Δt) = 0.1 s
Average force = -120 kg·m/s / (0.1 s)
Average force = -1200 N
Therefore, the average force player B must have exerted on player A during the collision is -1200 N (direction to the left).
c. Similarly, to calculate the average force player A must have exerted on player B during the collision, we can use Newton's second law of motion:
Average force = Change in momentum / Time
Given:
Change in momentum of player B = -Change in momentum of player A (due to conservation of momentum)
Change in momentum of player B = -(-120 kg·m/s) = +120 kg·m/s
Time (Δt) = 0.1 s
Average force = 120 kg·m/s / (0.1 s)
Average force = 1200 N
Therefore, the average force player A must have exerted on player B during the collision is +1200 N (direction to the right).
d. The change in momentum of player B can be calculated using the formula:
Change in momentum = mass of player B * (final velocity of player B - initial velocity of player B)
Given:
Mass of player B (mB) = 68 kg
Initial velocity of player B (uB) = -1.2 m/s (negative sign indicates direction to the left)
Final velocity of player B (vB) = ?
Change in momentum = 68 kg * (vB - (-1.2 m/s))
Change in momentum = 68 kg * (vB + 1.2 m/s)
Since momentum is conserved in a collision, the change in momentum of player B should also equal -120 kg·m/s. Therefore:
-120 kg·m/s = 68 kg * (vB + 1.2 m/s)
Solving for vB:
-120 kg·m/s = 68 kg * vB + 68 kg * 1.2 m/s
-120 kg·m/s - 68 kg * 1.2 m/s = 68 kg * vB
Dividing both sides by 68 kg:
(-120 kg·m/s - 68 kg * 1.2 m/s) / 68 kg = vB
vB ≈ -3.4 m/s
Therefore, the final velocity of player B is approximately -3.4 m/s (direction to the left).
Note: Since the velocity of player B is negative, it indicates the direction is to the left. In this context, the right side is considered positive and the left side is negative.
e. The final velocity of player B has already been calculated in the previous step, which is approximately -3.4 m/s.
original a momentum = 80*2.5
original b momentum = -68*1.2
total momentum before and after
= 80*.5 - 68*1.2
after
= 80*1 + 68 Vb
which is
final momentum a = 80
final momentum b = total - 80
F on a = (final momentum a - initial momentum a) /0.1