Question

A 25g golf ball moving with a speed of 40ms-1 lands in soft dirt and becomes embedded at a distance of 8 cm . Assuming constant acceleration during impact find the average force​

Answer 1

Given:

A 25g golf ball moving with a speed of 40ms-1 lands in soft dirt and becomes embedded at a distance of 8 cm .

To find:

Average force experienced by ball.

Calculation:

First of all , let's calculate the acceleration of the ball ;

$\rm{ {v}^{2} = {u}^{2} + 2ah}$

$= > \rm{ {(0)}^{2} = {(40)}^{2} + 2a( \frac{8}{100} )}$

$= > \rm{ 0 = 1600 + 0.16a}$

$= > \rm{ 0.16a = - 1600}$

$= > \rm{ a = - \dfrac{1600}{0.16} }$

$= > \rm{ a = - \dfrac{16 \times {10}^{4} }{16} }$

$= > \rm{ a = - {10}^{4} \: m {s}^{ - 2} }$

Now , average force be F ;

$\rm{ \therefore \:F = ma}$

$\rm{ = > \:F = \dfrac{25}{1000} \times ( - {10}^{4}) }$

$\rm{ = > \:F = 25 \times ( - 10) }$

$\rm{ = > \:F = - 250 \: newton}$

So, final answer is:

$\boxed{\bf{ \:F = - 250 \: newton}}$

• Negative sign of force signifies resistive force experienced by the ball after entering mud.

Answer 2

Answer:

Given:

A 25g golf ball moving with a speed of 40ms-1 lands in soft dirt and becomes embedded at a distance of 8 cm .

To find:

Average force experienced by ball.

Calculation:

First of all , let's calculate the acceleration of the ball ;

\rm{ {v}^{2} = {u}^{2} + 2ah}v

2

=u

2

+2ah

= > \rm{ {(0)}^{2} = {(40)}^{2} + 2a( \frac{8}{100} )}=>(0)

2

=(40)

2

+2a(

100

8

)

= > \rm{ 0 = 1600 + 0.16a}=>0=1600+0.16a

= > \rm{ 0.16a = - 1600}=>0.16a=−1600

= > \rm{ a = - \dfrac{1600}{0.16} }=>a=−

0.16

1600

= > \rm{ a = - \dfrac{16 \times {10}^{4} }{16} }=>a=−

16

16×10

4

= > \rm{ a = - {10}^{4} \: m {s}^{ - 2} }=>a=−10

4

ms

−2

Now , average force be F ;

\rm{ \therefore \:F = ma}∴F=ma

\rm{ = > \:F = \dfrac{25}{1000} \times ( - {10}^{4}) }=>F=

1000

25

×(−10

4

)

\rm{ = > \:F = 25 \times ( - 10) }=>F=25×(−10)

\rm{ = > \:F = - 250 \: newton}=>F=−250newton

So, final answer is:

\boxed{\bf{ \:F = - 250 \: newton}}

F=−250newton