Jogging in a park, an athlete runs at a speed of 6 km/h. He completes 4 rounds of the park daily.
If a complete round of the park is 200 m, find the time in minutes he takes to complete four rounds.
Answers
Answer:
The velocity acquired by the block is 4.4m/s.
Explanation:
Consider the provided information.
We've been given that a bullet of mass 10 g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900g.
With this information, we've been asked to find out the velocity acquired by the block.
Consider the mass of the bullet be m_1m
1
gram, mass of the wooden block be m_2m
2
gram, the initial velocity of the bullet be uu m/s and the final velocity acquired by the block be vv m/s.
Mass of the bullet, m_1 = 10g = \frac{10}{1000}kg = 0.01kgm
1
=10g=
1000
10
kg=0.01kg
Mass of the wooden block, m_2 = 900g = \frac{900}{1000}kg = 0.9kgm
2
=900g=
1000
900
kg=0.9kg
We can find the final velocity acquired by the block using the moment conservation of law.
According to the law of conservation of linear momentum, for an object or system of objects, the total momentum of the system is always conserved if no external force acts on them.
m_1 u_1 + m_2v_2 = (m_1 + m_2) \times vm
1
u
1
+m
2
v
2
=(m
1
+m
2
)×v
\begin{gathered}\implies v = \dfrac{m_1 u_1 + m_2v_2}{m_1 + m_2} \\ \\ \implies v = \dfrac{0.01 \times 400 + 0.9 \times 0}{0.01 + 0.9} \\ \\ \implies v = \dfrac{4+0}{0.91} \\ \\ \implies v = \dfrac{4}{0.91} \\ \\ \implies \boxed{v = 4.4}\end{gathered}
⟹v=
m
1
+m
2
m
1
u
1
+m
2
v
2
⟹v=
0.01+0.9
0.01×400+0.9×0
⟹v=
0.91
4+0
⟹v=
0.91
4
⟹
v=4.4
Hence, the final velocity acquired by the block is 4.4m/s.
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