Question

A 72.0 kg football player is gliding across very smooth ice at 2.25 m/s. He throws a 0.420 kg football straight forward. What is the combined velocity of the player and the football after the throw?

A) 1.0 m/s
B) 1.25 m/s
C) 1.5 m/s
D) 1.75 m/s

Answer 1

The combined velocity of the player and the football after the throw is 385.7 m/s.

None of the given options is correct

Step-by-step explanation:

To solve this problem, we can use the principle of conservation of momentum. The momentum before the throw is equal to the momentum after the throw. The momentum of an object is given by the product of its mass and velocity.

Let's define the velocity of the player before the throw as vp and the velocity of the football before the throw as vf.

Before the throw, the combined momentum of the player and the football is given by: mpvp + mfvf

After the throw, the combined momentum is given by: (mp + mf)v, where v is the combined velocity of the player and the football after the throw.

Since the player and the football are initially at rest, we can set the initial momentum to zero. This gives us the equation: 0 = mpvp + mfvf

Using the given values, we can substitute them into the equation and solve for v:

0 = (72.0 kg)(2.25 m/s) + (0.420 kg)v

Simplifying the equation:

0 = 162.0 kg·m/s + (0.420 kg)v

Solving for v:

(0.420 kg)v = -162.0 kg·m/s

v = -162.0 kg·m/s / 0.420 kg

v = -385.7 m/s

Since velocity is a vector quantity, we can disregard the negative sign and conclude that the combined velocity of the player and the football after the throw is 385.7 m/s.

None of the given options is correct