Answer:
B) The same as the momentum change of the heavier fragment.
Explanation:
Since the initial momentum of the system is zero, we have
0 = p + p' where p = momentum of lighter fragment = mv where m = mass of lighter fragment, v = velocity of lighter fragment, and p' = momentum of heavier fragment = m'v' where m = mass of heavier fragment = 25m and v = velocity of heavier fragment.
0 = p + p'
p = -p'
Since the initial momentum of each fragment is zero, the momentum change of lighter fragment Δp = final momentum - initial momentum = p - 0 = p
The momentum change of heavier fragment Δp' = final momentum - initial momentum = p' - 0 = p' - 0 = p'
Since p = -p' and Δp = p and Δp' = -p = p ⇒ Δp = Δp'
<u>So, the magnitude of the momentum change of the lighter fragment is the same as that of the heavier fragment. </u>
So, option B is the answer
Answer:
See the answers below
Explanation:
In order to solve these problems, we must decompose the magnitudes of the velocities on the x & y-axes, using the angles that are given. It is important for a better understanding to look at the attached image and understand how the angles are located relative to the horizontal axis.
1.
The horizontal component of the angle can be found with the cosine function. While the vertical component can be found using the sine function of the angle.
Viy = 24*sin (40)
Viy = 15.42 [m/s]
2.
The horizontal component of the angle can be found with the cosine function.
Vix = 24*cos(40)
Vix = 18.38 [m/s]
3.
Vix = 55*cos(25)
Vix = 49.84 [m/s]
The answer is C.) 10 m East
It is a synthesis reaction where two reactants gives one product.
Answer:option D
2Mg + O2 ------> 2 MgO
