Question

PLEASE GIVE BRANLIEST
Two hockey pucks with mass 0.1 kg slide across the ice and collide. Before the collision, puck 1 is going 13 m/s to the east and puck 2 is going 18 m/s to the west. After the collision, puck 1 is going 18 m/s to the west. What is the velocity of puck 2?

Answer 1

Explanation:

please mark me as the brainliest also

Answer:

13 m/s east

Explanation:

We can solve the problem by using the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision:

where

m = 0.1 kg is the mass of each puck

u1 = +13 m/s is the initial velocity of puck 1

u2 = -18 m/s is the initial velocity of puck 2 (here I assume the west direction to be the negative direction, so I put a negative sign)

v1 = -18 m/s is the final velocity of puck 1

v2 = ? is the final velocity of puck 2

Simplifying m from the formula and substituting the data, we can find the final velocity of puck 2, v2:

And the positive sign means that puck 2 is moving east.

Answer 2

We can solve the problem by using the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision:

\begin{gathered}p_i = p_f \\m u_1 + m u_2 = m v_1 + m v_2\end{gathered}

p i=p f

mu 1+mu 2=mv 1 +mv 2

where

m = 0.1 kg is the mass of each puck

u1 = +13 m/s is the initial velocity of puck 1

u2 = -18 m/s is the initial velocity of puck 2 (here I assume the west direction to be the negative direction, so I put a negative sign)

v1 = -18 m/s is the final velocity of puck 1

v2 = ? is the final velocity of puck 2

Simplifying m from the formula and substituting the data, we can find the final velocity of puck 2, v2:

v_2 = u_1 + u_2 - v_1 = +13 m/s + (-18 m/s) - (-18 m/s) = +13 m/sv 2

=u 1 +u 2 −v 1

=+13m/s+(−18m/s)−(−18m/s)=+13m/s

And the positive sign means that puck 2 is moving east.

13m/s.......

MY DEAR FRIEND PLEASE THANKS MY 15 ANSWERS ☺☺