Help Not sure what im doing wrong

[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.

 

[OlderDan]

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

 

[BunDa4Th]

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?

 

[OlderDan]

V is the velocity of the moving block.

That is not what it is here

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.

 

[BunDa4Th]

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

 

[PhanthomJay]

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??

 

[OlderDan]

m_bulletv_i = (m + M)g

Mass time velocity cannot possibly equal mass times acceleration.
You had this correct earlier

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

 

[BunDa4Th]

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

 

[PhanthomJay]

(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.

 

[BunDa4Th]

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

which is t = -V/6.37

 

[PhanthomJay]

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.

 

[BunDa4Th]

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.