A young hockey player stands at rest on the ice holding a 1.3-kg helmet. The player tosses the helmet with a speed of 4.6 m/s in a direction 13° above the horizontal, and recoils with a speed of 0.27 m/s. Find the mass of the hockey player.

To find the mass of the hockey player, we can use the principle of conservation of momentum.

1. The initial momentum of the system, which includes the hockey player and the helmet, is zero since the player is at rest.
2. The final momentum of the system is also zero since the player recoils with a speed of 0.27 m/s, which is opposite in direction to the initial momentum.
3. The momentum of an object is calculated by multiplying its mass by its velocity. In this case, the mass of the helmet is given as 1.3 kg.

Let's denote the mass of the hockey player as "m."

The horizontal component of the helmet's momentum can be calculated using the formula: p = mv.

The initial horizontal component of the helmet's momentum is given by:
p_initial = m * 4.6 m/s * cos(13°)

The final horizontal component of the helmet's momentum is given by:
p_final = m * (-0.27 m/s)

Since the initial and final momentums are equal (zero in this case), we can write the following equation:

m * 4.6 m/s * cos(13°) = m * (-0.27 m/s)

Now, we can solve this equation to find the mass of the hockey player.

4.6 m/s * cos(13°) = -0.27 m/s (dividing both sides by m)
4.6 m/s * cos(13°) / (-0.27 m/s) = m

Using a calculator, we can evaluate the left side of the equation:

m ≈ -99.674 kg

However, mass cannot be negative, so we discard this result.

Therefore, there is an error in the problem statement or the calculations provided. Please double-check the given information or calculations to find the correct result.

To solve this problem, we need to use the principle of conservation of momentum. According to this principle, the total momentum before the helmet is thrown should be equal to the total momentum after the helmet is thrown.

The momentum of an object is given by the product of its mass and velocity. So, before the helmet is thrown, the total momentum is the momentum of the hockey player holding the helmet:

momentum_before = mass_of_hockey_player * velocity_of_hockey_player

After the helmet is thrown, the momentum of the hockey player and the helmet can be calculated separately.

First, let's find the momentum of the helmet. The momentum of the helmet is given by the product of its mass and velocity:

momentum_of_helmet = mass_of_helmet * velocity_of_helmet

Next, let's find the momentum of the hockey player after the helmet is thrown. The hockey player recoils with a velocity of 0.27 m/s:

momentum_of_hockey_player = mass_of_hockey_player * velocity_of_hockey_player_after

Since the total momentum before and after the helmet is thrown should be the same, we can write:

momentum_before = momentum_of_helmet + momentum_of_hockey_player

mass_of_hockey_player * velocity_of_hockey_player = mass_of_helmet * velocity_of_helmet + mass_of_hockey_player * velocity_of_hockey_player_after

We can rearrange this equation to solve for the mass of the hockey player:

mass_of_hockey_player = (mass_of_helmet * velocity_of_helmet) / (velocity_of_hockey_player - velocity_of_hockey_player_after)

Plugging in the given values:

mass_of_helmet = 1.3 kg
velocity_of_helmet = 4.6 m/s
velocity_of_hockey_player = 0 m/s (since the player is at rest)
velocity_of_hockey_player_after = 0.27 m/s

mass_of_hockey_player = (1.3 kg * 4.6 m/s) / (0 m/s - 0.27 m/s)

mass_of_hockey_player = -6.42 kg/m/s

Therefore, the mass of the hockey player is -6.42 kg/m/s.

Its another momentum problem similar to your other post. Since there is only two 'fragments' from the 'explosion' :P you can disregard the angle for the velocity and act like they are horizantal to each other. If you don't want to disregard the angle, then you may use component method and find the two component momentums then use pythagorean theorom.

Easier way is to just impose they're horizantal, therefore:
m1v1=-m2v2 (since they are going in different directions, the momentums will be the negative of each other).
You have two velocities and a amss, sub for the final mass.