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Answer :
Final answer:
The resulting velocity of the 77.8 kg basketball player after the collision is 4.48 m/s.
Explanation:
To find the resulting velocity of the 77.8 kg basketball player after the collision, we can use the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision.
Before the collision, the momentum of the basketball player is given by mass times velocity: (77.8 kg)(8.1 m/s) = 628.18 kg·m/s. The momentum of the 99.8 kg player is (99.8 kg)(-6.9 m/s) = -687.42 kg·m/s, as it is moving in the opposite direction.
After the collision, the momentums must add up to zero since they cancel each other out. Let's call the resulting velocity of the basketball player v:
(77.8 kg)(v) + (99.8 kg)(-3.5 m/s) = 0
Solving for v, we get v = (99.8 kg)(3.5 m/s) / 77.8 kg = 4.48 m/s. Therefore, the resulting velocity of the 77.8 kg basketball player is 4.48 m/s.
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Answer:
-5.24 m/s
** The minus sign indicates that the velocity vector points in the opposite direction with respect to the initial direction of the 77.8 kg player **
Explanation:
Hi!
We can solve this problem considering each player as a point particle and taking into account the conservation of linear momentum.
Since the 99.8 kg player is moving towards the 77.8kg, the initial total momentum is:
m1*v1_i + m2*v2_i = (77.8kg)(8.1 m/s) - (99.8kg)(6.9 m/s)
** The minus sign indicates that the velocity vector points in the opposite direction with respect to the initial direction of the 77.8 kg player **
The final total momentum is equal to:
m1*v1_f + m2*v2_f = (77.8 kg)v1_f + (99.8 kg)(3.5 m/s)
The conservation of momentu tell us that:
m1v1_i + m2v2_i = m1v1_f + m2v2_f
Therefore:
v1_f =v1_i + (m2/m1)*(v2_i-v2_f)
v1_f = 8.1 m/s + (99.8 / 77.8) * (-6.9 - 3.5 m/s)
v1_f = -5.24 m/s
