Question

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.

Answer 1

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

Answer 2

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