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mash [69]
2 years ago
10

Two rocks of masses 5 kg and 10 kg a dropped from rest from a height of 20 m above ground. Ignore all friction and air resistanc

e. Just before
they hit the ground, which statement is true about the rocks?
(A) The value for the acceleration of the 5 kg rock is the same as the value for the 10 kg rock, but their velocity values are different.
(B) The values for the acceleration and velocity of the 5 kg rock are greater than the values for the 10 kg rock.
(C) The values for the acceleration and velocity of the 5 kg rock are less than the values for the 10 kg rock.
(D) The values for the acceleration and velocity of the 5 kg rock are the same as the values for the 10 kg rock.
Physics
1 answer:
kupik [55]2 years ago
6 0

Answer:

D

Explanation:

F=ma

F of 5 kg rock = 49N

F of 10 kg rock = 98 N

Divide by respective masses to get acceleration, and of course you will get 9.8 m/s^2 for both.

Now, use potential energy equals kinetic energy. mgh=(1/2)mv^2 mass cancels out of the equation, since it's on both sides, so we can stop right there. We have algebraically determined that mass does not affect acceleration or velocity!

Hope this helped.

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malfutka [58]

Answer:

a) p_i=1.568\hat{i}+0.752 \hat{j}

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c) v_{fy}=0.7999\ m.s^{-1}

Explanation:

Given masses:

m_1=0.49\ kg

m_2=0.47\ kg

Velocity of mass 1, v_1=3.2 \hat{i}\ m.s^{-1}

Velocity of mass 2, v_2=1.6 \hat{j}\ m.s^{-1}

a)

Initial momentum:

p_i=m_1.v_1+m_2.v_2

p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}

p_i=1.568\hat{i}+0.752 \hat{j}

b)

magnitude of initial momentum:

p_i=\sqrt{1.568^2+0.752 ^2}

p_i=1.739\ kg.m.s^{-1}

From the conservation of momentum:

p_f=p_i

m_f.v_f=1.739

v_f=\frac{1.739}{0.49+0.47}

v_f=1.85\ m.s^{-1} is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

tan\theta=\frac{0.752 }{1.568}

\theta=25.62^{\circ}

\therefore v_{fx}=1.85\ cos25.62^{\circ}

v_{fx}=1.668\ m.s^{-1}

c)

Vertical component of final velocity:

v_{fy}=1.85\ sin 25.62^{\circ}

v_{fy}=0.7999\ m.s^{-1}

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