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OlgaM077 [116]
3 years ago
15

The collision between a hammer and a nail can be considered to be approximately elastic. estimate the kinetic energy acquired by

a 10 g nail when it is struck by a 550 g hammer moving with an initial speed of 3.5 m/s.
Physics
2 answers:
Setler [38]3 years ago
6 0

Here we can use momentum conservation as in this type of collision there is no external force on it

m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}

now here we can say

m_1 = 10 g

v_{1i} = 0

m_2 = 550 g

v_{2i} = 3.5 m/s

now here we can say

10*0 + 550 * 3.5 = 10 v_{1f} + 550 v_{2f}

192.5 = v_{1f} + 55 v_{2f}

now by coefficient of restitution

for elastic collision we know that e = 1

v_{2f} - v_{1f} = e(v_{1i} - v_{2i})

v_{2f} - v_{1f} = 0 - 3.5

now by solving the two equation

56v_{2f} = 189

v_{2f} = 3.375 m/s

also we know that

v_{1f} = v_{2f} + 3.5 = 3.375 + 3.5 = 6.875 m/s

so final speed of the nail is 6.875 m/s


belka [17]3 years ago
6 0

Answer:

0.236J

Explanation:

Given:  collision between a hammer and a nail, it is approximately elastic.a 550 g hammer moving with an initial speed of 3.5 m/s struck a 10g nail.

To Find: Kinetic Energy acquired by nail.

Solution: Let mass and initial speed of hammer be=\text{m}_{1} , \text{v}_{1i}

final speed of hammer=\text{v}_{1f}

                mass and initial speed of nail be=\text{m}_{2} , \text{v}_{2i}

                final speed of nail=\text{v}_{2f}

momentum before collision

\text{m}_{1} \text{v}_{1i}  + \text{m}_{2} \text{v}_{2i}

momentum after collision

\text{m}_{1} \text{v}_{1f}  + \text{m}_{2} \text{v}_{2f}

as collision is elastic momentum is conserved

momentum before collision = momentum after collision

\text{m}_{1} \text{v}_{1i} + \text{m}_{2} \text{v}_{2i} = \text{m}_{1} \text{v}_{1f} + \text{m}_{2} \text{v}_{2f}

as nail was at rest initially , \text{v}_{2i} = 0

\text{m}_{1} v_{1i}=\text{m}_{1} v_{1f} +\text{m}_{2} v_{2f}

\text{m}_{1}(\text{v}_{1i} -\text{v}_{1f}) = m_{2} v_{2f}

\frac{\text{m}_{1}}{\text{m}_{2}} = \frac{\text{v}_{2f}^2}{\text{v}_{1i}-\text{v}_{1f}}

kinetic energy before collision

\frac{1}{2}\text{m}_{1}\text{v}_{1i}^{2} +\frac{1}{2}\text{m}_{2}\text{v}_{2i}^{2}

kinetic energy after collision

\frac{1}{2}\text{m}_{1}\text{v}_{1f}^{2} +\frac{1}{2}\text{m}_{2}\text{v}_{2f}^{2}

As in elastic collision Kinetic energy remains conserved

kinetic energy before collision= kinetic energy after collision

\frac{1}{2}\text{m}_{1}\text{v}_{1i}^{2} +\frac{1}{2}\text{m}_{2}\text{v}_{2i}^{2} = \frac{1}{2}\text{m}_{1}\text{v}_{1f}^{2} +\frac{1}{2}\text{m}_{2}\text{v}_{2f}^{2}

given, v_{2i} = 0

\text{m}_{1}(\text{v}_{1i}^2-\text{v}_{1f}^2) = \text{m}_{2}\text{v}_{2f}^2

\frac{\text{m}_{1}}{\text{m}_{2}}= \frac{\text{v}_{2f}^{2}}{(\text{v}_{1i}-\text{v}_{1f})^{2}}

putting value of \frac{\text{m}_{1}}{\text{m}_{2}} from previous equation

\frac{\text{v}_{2f}}{\text{v}_{1i}-\text{v}_{1f}}=\frac{\text{v}_{2f}^{2}}{(\text{v}_{1i}-\text{v}_{1f})^{2}}

\text{v}_{2f} = \text{v}_{1i} + \text{v}_{1f}

putting it in equation of momentum, we get

\frac{\text{v}_{1i}}{\text{v}_{1f}}=\frac{\text{m}_{1}+\text{m}_{2}}{\text{m}_{1}-\text{m}_{2}}

putting values \text{v}_{1f}= 3.375\text{m}\setminus\text{s}

\text{v}_{2f} = \text{v}_{1i} + \text{v}_{1f}

\text{v}_{2f} = \text{3.5} + \text{3.375}

\text{v}_{2f} = 6.875

Kinetic energy acquired by nail =\frac{1}{2}\text{m}\text{v}_{2f}^2

                                                     \frac{1}{2}\times 0.01\times 6.875^2

                                                      0.236 J

Hence Kinetic Energy acquired by nail is 0.236 J  

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