Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.
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A 100kg crate slides along a floor with a starting velocity of 21 m/s. If the force due to friction is 8N, then, it will take 262.5 s for the box to come to rest.
We'll begin by calculating the declaration of the box. This can be obtained as follow:
Force (F) = –8 N (opposition)
Mass (m) = 100 Kg
<h3>Deceleration (a) =? </h3>
<h3>F = ma</h3>
–8 = 100 × a
Divide both side by 1000

<h3>a = –0.08 ms¯²</h3>
Therefore, the deceleration of the box is –0.08 ms¯²
Finally, we shall determine the time taken for the box to come to rest. This can be obtained as follow:
Deceleration (a) = –0.08 ms¯²
Initial velocity (u) = 21 ms¯¹
Final velocity (v) = 0 ms¯¹
<h3>Time (t) =.? </h3>
<h3>v = u + at</h3>
0 = 21 + (–0.08×t)
0 = 21 – 0.08t
Collect like terms
0 – 21 = –0.08t
–21 = –0.08t
Divide both side by –0.08

<h3>t = 262.5 s</h3>
Therefore, it will take 262.5 s for the box to come to rest.
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Answer:
v_g,i = 1.208 m/s
Explanation:
We are given;
Mass of girl; m_g = 47.2 kg
Mass of plank; m_p = 177 kg
Let the velocity of girl to ice be v_g,i
Let the velocity of plank to ice be v_p,i
Since the velocity of the girl is 1.53 m/s relative to the plank, then;
v_g,i + v_p,i = 1.53
From conservation of momentum;
m_g × v_g,i = m_p × v_p,i
Thus;
47.2(v_g,i) = 177(v_p,i)
Dividing both sides by 47.2 gives;
v_g,i = 3.75(v_p,i)
v_pi = (v_g,i)/3.75
Thus, from v_g,i + v_p,i = 1.53, we have;
v_g,i + ((v_g,i)/3.75) = 1.53
v_g,i(1 + 1/3.75) = 1.53
1.267v_g,i = 1.53
v_g,i = 1.53/1.267
v_g,i = 1.208 m/s
Same magnitude (1000N) and opposite direction
you may also answer -1000 N