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

The kinetic energy acquired by the 10 g nail when struck by a 550 g hammer moving with an initial speed of 3.5 m/s is approximately 0.0275 Joules.

Step-by-step explanation:

In an approximately elastic collision, both momentum and kinetic energy are conserved. To estimate the kinetic energy acquired by the nail, we can use the equation for kinetic energy, which is given by:

Kinetic energy (KE) = 0.5 * mass * velocity^2

The mass of the nail is 10 g, which is equal to 0.01 kg. The initial velocity of the hammer is 3.5 m/s. Plugging these values into the equation:

KE = 0.5 * 0.01 kg * (3.5 m/s)^2

KE = 0.0275 Joules

Therefore, the kinetic energy acquired by the 10 g nail is approximately 0.0275 Joules.

Answer 2

0.236J

Step-by-step 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

`(1)/(2)\text{m}_(1)\text{v}_(1i)^(2) +(1)/(2)\text{m}_(2)\text{v}_(2i)^(2)`

kinetic energy after collision

`(1)/(2)\text{m}_(1)\text{v}_(1f)^(2) +(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

`(1)/(2)\text{m}_(1)\text{v}_(1i)^(2) +(1)/(2)\text{m}_(2)\text{v}_(2i)^(2)` =
`(1)/(2)\text{m}_(1)\text{v}_(1f)^(2) +(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 =
`(1)/(2)\text{m}\text{v}_(2f)^2`

`(1)/(2)* 0.01* 6.875^2`

0.236 J

Hence Kinetic Energy acquired by nail is 0.236 J