Question

An urban vehicle weighing 2500 kg hits a stationary 1100kg vehicle. The vehicles stick together and continue to slide 50m to a stop. coefficient of friction = .7 . How fast is the EUV going before the collision

Answer 1

\[m_1v_1+m_2v_2=(m_1+m_2)v'\]

Answer 2

\[m_1v_1=(m_1+m_2)v'\]

Answer 3

\[v_1=\frac{(m_1+m_2)v'}{m_1}\]

Answer 4

don't know v' so we must find it

Answer 5

given 50m

\[50=\frac{1}{2}at^2 +v_0t+s_0\]
\[50=\frac{1}{2}at^2 +v_0t+s_0\]

\[v_f=at+v_0\]
since we kno wthat final velocity after collision is 0 (stops)
\[0=at+v_0 \]
\[-at=v_0\]

Answer 6

don't have time soooo maybe use the time independent equation

Answer 7

a= .7*3600*g

Vf^2 =0 = Vi^2 - 2ad
Vi^2 = 2(.7*3600*g)*50

Answer 8

where'd your acceleration come from?

Answer 9

\[0=v_0^2+2as\]
\[-2ag=v_0^2\]
\[\sqrt{-2ag}=v_0\]

Answer 10

\[\sqrt{-2as}=v_0\]

Answer 11

so now we can replace this with v_0 and we can use the other equation to get rid of the time

Answer 12

none of that is necessary.

Answer 13

I showed you how to find the velocity immediately after the collision. Use the conservation of momentum equation to find initial velocity.

Answer 14

no you simply stated that the acceleration is the (friction coefficient) times the mass of the object

Answer 15

was this found out through a fbd?

Answer 16

whoops, yeah... a= .7g... shouldn't have mass in there...

Answer 17

a= .7*g

Vf^2 =0 = Vi^2 - 2ad
Vi^2 = 2(.7*g)*50

Answer 18

mu*N is the force ma = mu*N = mu*mg
-> a=mu*g

Answer 19

\[\sum F_x=ma=-f\]

\[F_y=ma=0=N-W=\]
\[N=W\]
\[N=mg\]
\[\mu N=.7*mg\]

Answer 20

alright i wasn't sure how the 50 meters factored in?

Answer 21

a= .7*g

Vf^2 =0 = Vi^2 - 2ad
Vi^2 = 2(.7*g)*50

Answer 22

yeah i thought it was to find acceleration alright thanks

Answer 23

sure