# Help Not sure what im doing wrong

• BunDa4Th
In summary: Mv_2 = (m + M)V11.4v_i = (11.4 + 94)(9.8)11.4v_i = 1032.92v_i = 90.61 m/sMomentum is conserved at impact, but why do you say m_bulletv_i = (m +M)g? where does the g come from??m_bulletv_i = (m + M)g Mass time velocity cannot possibly equal mass times acceleration.
BunDa4Th
Help! Not sure what I am doing wrong

A 11.4 g bullet is fired horizontally into a 94 g wooden block initially at rest on a horizontal surface. After impact, the block slides 7.5 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.650, what was the speed of the bullet immediately before impact? m/s

first thing i did was find the block speed

t = sqrt((7.5)(.65)/9.8)

t = .7053

then i did

V = deltax/t = 7.5/.7053 = 10.6338

mv_1 + Mv_2 = (m + M)V

V = 98.31 which is incorrect but answer is within 10% of the correct value.

I am not sure if I am doing this correctly or I am just completely off.

BunDa4Th said:
A 11.4 g bullet is fired horizontally into a 94 g wooden block initially at rest on a horizontal surface. After impact, the block slides 7.5 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.650, what was the speed of the bullet immediately before impact? m/s

first thing i did was find the block speed

t = sqrt((7.5)(.65)/9.8)

t = .7053

then i did

V = deltax/t = 7.5/.7053 = 10.6338

mv_1 + Mv_2 = (m + M)V

V = 98.31 which is incorrect but answer is within 10% of the correct value.

I am not sure if I am doing this correctly or I am just completely off.
Your V is on the order of 10% too high. It should be less than 10m/s

t = sqrt(mu*s/g)?

I assume that last V means v_1

Last edited:
V is the velocity of the moving block.

Also i thought to get "t" it is t = sqrt(deltaX/g) but i was not sure how to get it on here since friction play a part of this problem.

t = sqrt(mu*s/g) what is "s"? i know that mu is the kinetic friction and g is gravity but what is s?

BunDa4Th said:
V is the velocity of the moving block.
That is not what it is here
BunDa4Th said:
V = 98.31 which is incorrect but answer is within 10% of the correct value.
Also i thought to get "t" it is t = sqrt(deltaX/g) but i was not sure how to get it on here since friction play a part of this problem.

t = sqrt(mu*s/g) what is "s"? i know that mu is the kinetic friction and g is gravity but what is s?
s is often used for distance. You called it deltaX here. So change my query to

t = sqrt(mu*deltaX/g)? There is no such equation.

From V_f - V = a*t you get V = -a*t (zero final velocity)
V is the initial velocity of the block after collision, as you defined it. The acceleration is negative (decleration). So

t = -V/a

apply Newton's second law to find a

(M+m)a = F_f = force of friction = mu*N = mu*(M+m)g

etc

Alternatively, once you identify a from the above, instead of finding the time you can use the equation that relates the change of velocity squared to acceleration and distance moved.

okay you just completely lost me there.

but i seem to got the answer somehow by doing this way which is probably incorrect to do it and i just got lucky and got it correct (this is what i got from reading what you posted). I am trying to figure out what you are telling me but i can't seem to understand it. sorry if I am giving you a much harder time on this, but my professor never went over this (she skip more than half of the chapter) and expect us to know it.

m_bulletv_i = (m + M)g

11.4v_i = (11.4 + 94)(9.8)

11.4v_i = 1032.92
v_i = 90.61 m/s

BunDa4Th said:
okay you just completely lost me there.

but i seem to got the answer somehow by doing this way which is probably incorrect to do it and i just got lucky and got it correct (this is what i got from reading what you posted). I am trying to figure out what you are telling me but i can't seem to understand it. sorry if I am giving you a much harder time on this, but my professor never went over this (she skip more than half of the chapter) and expect us to know it.

m_bulletv_i = (m + M)g

11.4v_i = (11.4 + 94)(9.8)

11.4v_i = 1032.92
v_i = 90.61 m/s
Momentum is conserved at impact, but why do you say m_bulletv_i = (m +M)g? where does the g come from??

BunDa4Th said:
m_bulletv_i = (m + M)g
Mass time velocity cannot possibly equal mass times acceleration.
mv_1 + Mv_2 = (m + M)V
Just recognize that v_2= 0

Solve the equation I started for you relating (M+m)a to mu*(M+m)*g to find a. Find the value of V from the known stopping distance and the acceleration. Then use V in the equation above to find v_1

(94 + 11.4)a = .65(94 + 11.4) (9.8)
105.4a = 671.398
a = 6.37 is this correct?

BunDa4Th said:
(94 + 11.4)a = .65(94 + 11.4) (9.8)
105.4a = 671.398
a = 6.37 is this correct?
Yes, now proceed per Dan's earlier responses.

that is where i am stuck how would i go from there if t = -V/a

which is t = -V/6.37

BunDa4Th said:
that is where i am stuck how would i go from there if t = -V/a

which is t = -V/6.37
It's best now that you know what a is, to use v^2 = 2as, wher s = 7.5 meters. and a = 6.37m/s/s. This is the initial velocity of the block/bullet system after impact. Then use conservation of momentum to solve for v_bullet. Make sure to include units in your answers.

Thanks a lot it was correct. Thanks for taking the time on explaining and giving a step by step on how to do this. I forgot about the kinetic equation. Both of you were great help on explaining and taking the time to help me solve this.

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