Question

Two ice skaters collide on the ice. A 39.6-kg skater moving South at 6.21 m/s collides with a 52.1-kg skater moving East at 4.33 m/s. The two skaters entangle and move together across the ice. Determine the magnitude and direction of their post-collision velocity.

Answer 1

Answer:

V = 3.6385 m/s

θ = 47.46 degrees

Explanation:

the important data in the question is:

Skater 1:

$M_1$= 39.6 kg

direction: south (axis y)

$V_{1iy}$ = 6.21 m/s

Skater 2:

$M_2$ = 52.1 kg

direction: east (axis x)

$V_{2ix}$ = 4.33 m/s

Now using the law of the conservation of linear momentum ( $P_i = P_f$ and knowing that the collision is inelastic we can do the next equations:

$M_{1}V_{1ix}+M_2V_{2ix} = V_{sx}(M_1+M_2)$  (eq. 1)

$M_{1}V_{1iy}+M_2V_{2iy} = V_{sy}(M_1+M_2)$  (eq. 2)

Where $V_{sx}$ and $V_{sy}$ is the velocity of the sistem in x and y after the collision.

Note: the conservation of the linear momentum have to be make once by each axis.

Now, in the (eq. 1) the skater 1 don't have velocity in the axis x, so we can replace $V_{1ix}$ by 0 in the equation and get:

$M_2V_{2ix} = V_{sx}(M_1+M_2)$  (eq. 1)

also, in the (eq. 2) the skater 2 don't have velocity in the axis y, so we can replace $V_{2iy}$ by 0 in the equation and get:

$M_{1}V_{1iy} = V_{sy}(M_1+M_2)$  (eq. 2)

Now, we just replace the data in both equations:

$(52.1)(4.33) = V_{sx}(39.6+52.1)$  (eq. 1)

$(39.6)(6.21) = V_{sy}(39.6+52.1)$  (eq. 2)

solving for $V_{sx]$ and $V_{sy}$ we have:

$V_{sx]$ = 2.46 m/s

$V_{sy]$ = 2.681 m/s

using the pythagoras theorem we can find the magnitude of the velocity as:

V = $\sqrt{2.46^2+2.681^2}$

V = 3.6385 m/s

For find the direction we just need to do this;

θ = $tan^{-1}(\frac{2.681}{1.46})$

θ = 47.46 degrees