ok so I really don't get 2d collisions and I thought I knew how to do this one but I have no idea:
A 104 kg baseball player jumps up verticlly to catch a baseball of mass .254 kg traveling horizontally with a speed of 28.6 m/s. If the vertical speed of the player is .258 m/s the instant he catches the ball, detemine him speed just after the catch.
please help or let me know how on Earth I would solve this!
duplicate
To solve this problem, you can use 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.
Firstly, let's calculate the initial momentum before the catch. The initial horizontal momentum is given by the product of the mass (0.254 kg) and the initial horizontal velocity (28.6 m/s), which results in:
Initial horizontal momentum = (0.254 kg) x (28.6 m/s) = 7.2744 kg·m/s
Since the baseball player is initially at rest horizontally, his initial horizontal momentum is zero.
Now, let's find the final vertical momentum after the catch. The final vertical momentum is given by the product of the mass (0.254 kg) and the final vertical velocity (0.258 m/s), which becomes:
Final vertical momentum = (0.254 kg) x (0.258 m/s) = 0.065732 kg·m/s
Since there is no vertical motion before the catch, the initial vertical momentum is zero.
According to the conservation of momentum, the total initial momentum is equal to the total final momentum. Therefore, the sum of the initial horizontal momentum and the initial vertical momentum is equal to the sum of the final horizontal momentum and the final vertical momentum:
Initial horizontal momentum + Initial vertical momentum = Final horizontal momentum + Final vertical momentum
0 + 0 = Final horizontal momentum + 0.065732 kg·m/s
Simplifying the equation, we find the final horizontal momentum alone is equal to:
Final horizontal momentum = -0.065732 kg·m/s
Since the baseball player catches the ball while jumping vertically, and assuming there are no external horizontal forces acting on him, the horizontal momentum after the catch remains constant.
Therefore, the player's speed just after the catch, which is the final horizontal velocity, is equal to the final horizontal momentum divided by the player's mass:
Player's speed just after the catch = |Final horizontal momentum| / Player's mass
Player's speed just after the catch = |-0.065732 kg·m/s| / 104 kg
Player's speed just after the catch ≈ 0.000631 m/s
So, the player's speed just after the catch is approximately 0.000631 m/s.