laf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400kg that is traveling horizontally at 11.5m/s . Olaf's mass is 65.2kg.
If the ball hits Olaf and bounces off his chest horizontally at 7.00m/s in the opposite direction, what is his speed vf after the collision?
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mastering physics gave answer
To solve this problem, we need to apply the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Momentum (p) is defined as the product of mass (m) and velocity (v): p = m * v.
Given:
Mass of the ball (m1) = 0.400 kg
Initial velocity of the ball (v1) = 11.5 m/s (horizontally)
Mass of Olaf (m2) = 65.2 kg
Initial velocity of Olaf (v2) = 0 m/s (since he is standing on the ice)
Let's calculate the total initial momentum:
Initial momentum (Pi) = (m1 * v1) + (m2 * v2)
= (0.400 kg * 11.5 m/s) + (65.2 kg * 0 m/s)
= 4.60 kg·m/s + 0 kg·m/s
= 4.60 kg·m/s
According to the principle of conservation of momentum, the total momentum after the collision (Pf) should be equal to the initial momentum (Pi):
Pf = Pi
Now, let's find the final momentum after the collision. The ball's mass and velocity remain the same, but the direction of the velocity is opposite, so we can write it as -7.00 m/s (since it's horizontally).
Final momentum (Pf) = (m1 * v1) + (m2 * vf)
= (0.400 kg * -7.00 m/s) + (65.2 kg * vf)
Since Pf = Pi, we can equate the two equations and solve for vf:
(0.400 kg * -7.00 m/s) + (65.2 kg * vf) = 4.60 kg·m/s
Now, let's solve for vf:
-2.80 kg·m/s + 65.2 kg·vf = 4.60 kg·m/s
65.2 kg·vf = 7.40 kg·m/s
vf = 7.40 kg·m/s / 65.2 kg
vf = 0.113 m/s
Therefore, Olaf's speed (vf) after the collision is approximately 0.113 m/s in the opposite direction.