# Momentum A2 Question

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A cricket ball of mass 0.16 kg travelling at a speed of 35 m s–1 is hit by a bat and, as aresult of the impact, leaves the bat in the opposite direction at 30 m s–1. If the durationof the impact is 52 ms, what is the magnitude of the average force on the ball?

A 0.015N B 0.20N C 15N D 200N

I got the answer as 15N, but it says the answer is 200N?

Please may I have a solution? Cheers!

A 0.015N B 0.20N C 15N D 200N

I got the answer as 15N, but it says the answer is 200N?

Please may I have a solution? Cheers!

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#2

F(delta)t = (delta)mv

m is constant so F(delta)t = m(delta)v

(delta)v = -30 - 35 = -65

So m(delta)v = 0.16*-65 = -10.4

Divide this by 0.052 to get F

Magnitude is asked for so you can discount the minus sign in the final answer

m is constant so F(delta)t = m(delta)v

(delta)v = -30 - 35 = -65

So m(delta)v = 0.16*-65 = -10.4

Divide this by 0.052 to get F

Magnitude is asked for so you can discount the minus sign in the final answer

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#3

(Original post by

A cricket ball of mass 0.16 kg travelling at a speed of 35 m s–1 is hit by a bat and, as aresult of the impact, leaves the bat in the opposite direction at 30 m s–1. If the durationof the impact is 52 ms, what is the magnitude of the average force on the ball?

A 0.015N B 0.20N C 15N D 200N

I got the answer as 15N, but it says the answer is 200N?

Please may I have a solution? Cheers!

**Sasman96**)A cricket ball of mass 0.16 kg travelling at a speed of 35 m s–1 is hit by a bat and, as aresult of the impact, leaves the bat in the opposite direction at 30 m s–1. If the durationof the impact is 52 ms, what is the magnitude of the average force on the ball?

A 0.015N B 0.20N C 15N D 200N

I got the answer as 15N, but it says the answer is 200N?

Please may I have a solution? Cheers!

Please post the working telling us how you arrived at your answer.

We can then point you in the right direction for you to arrive at the solution under your own steam and hence for YOU to understand where it went wrong.

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#4

(Original post by

F(delta)t = (delta)mv

m is constant so F(delta)t = m(delta)v

(delta)v = -30 - 35 = -65

So m(delta)v = 0.16*-65 = -10.4

Divide this by 0.052 to get F

**jf1994**)F(delta)t = (delta)mv

m is constant so F(delta)t = m(delta)v

(delta)v = -30 - 35 = -65

So m(delta)v = 0.16*-65 = -10.4

Divide this by 0.052 to get F

Thss does not help the OP learn and all you have done is completed his homework for him.

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#5

p=mv-mu

p=0.16*(-30)-0.16*35 ---> minus sign because velocity is a vector and its traveling the opposite direction to which it came from

F=ma=∆p/∆T ---->∆p/∆T=m(V2-V1)/∆T=m*(dv/dt)=ma (curly equal signs)

p=0.16*(-30)-0.16*35 ---> minus sign because velocity is a vector and its traveling the opposite direction to which it came from

F=ma=∆p/∆T ---->∆p/∆T=m(V2-V1)/∆T=m*(dv/dt)=ma (curly equal signs)

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(Original post by

F(delta)t = (delta)mv

m is constant so F(delta)t = m(delta)v

(delta)v = -30 - 35 = -65

So m(delta)v = 0.16*-65 = -10.4

Divide this by 0.052 to get F

Magnitude is asked for so you can discount the minus sign in the final answer

**jf1994**)F(delta)t = (delta)mv

m is constant so F(delta)t = m(delta)v

(delta)v = -30 - 35 = -65

So m(delta)v = 0.16*-65 = -10.4

Divide this by 0.052 to get F

Magnitude is asked for so you can discount the minus sign in the final answer

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#7

(Original post by

Please do not post full solutions to problems.

Thss does not help the OP learn and all you have done is completed his homework for him.

**uberteknik**)Please do not post full solutions to problems.

Thss does not help the OP learn and all you have done is completed his homework for him.

But I'll remember for next time...

e/ anytime Sasman

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#8

Why do you add both the incoming speed and final speed together please? thanks :-)

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#9

(Original post by

Why do you add both the incoming speed and final speed together please? thanks :-)

**JHousley**)Why do you add both the incoming speed and final speed together please? thanks :-)

so (0.16x35)-(0.16x-35)=10.4NS

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