# Momentum A2 Question

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#1
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!
0
5 years ago
#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

1
5 years ago
#3
(Original post by 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!
Hello and welcome to TSR physics!

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.
0
5 years ago
#4
(Original post by 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
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.
0
5 years ago
#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)
0
#6
(Original post by 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

Cheers mate, I was calculating it as (35x0.16)-(30x0.16)/52x10^-3 therefore getting 15N.
0
5 years ago
#7
(Original post by 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.
It's not homework, it's taken from one of the AQA past papers, I know because I recognise it, and the answer is already given

But I'll remember for next time...

e/ anytime Sasman
0
5 years ago
#8
Why do you add both the incoming speed and final speed together please? thanks :-)
0
5 years ago
#9
(Original post by JHousley)
Why do you add both the incoming speed and final speed together please? thanks :-)
depend which diretion u take positive so impulse=mv-mu if u took one direction to be positive one of the velocity has to ne negative as it is travelling in the opposite direction.
so (0.16x35)-(0.16x-35)=10.4NS
0
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