Answer :
Answer:
a) vs² = (ve + vn)²/4
b) K = m*(ve + vn)²/4
c) Wfric = - 79.461*m
d) vn = 13.2128 m/s < 14 m/s
Then, the driver N has reported his speed correctly.
Explanation:
Part A.
Given
me = mn = m
ve
vn
vsystem = vs = ?
We can apply the equation
me*ve + mn*vn = (me + mn)*vs
⇒ m*ve + m*vn = (m + m)*vs ⇒ m*(ve + vn) = 2m*vs
⇒ vs = (ve + vn)/2 ⇒ vs² = (ve + vn)²/4
Part B.
K = ?
We use the formula
K = (2m)*vs²/2 ⇒ K = (2m)*((ve + vn)²/4)/2
⇒ K = m*(ve + vn)²/4
Part C.
Given
μ = 0.9
d = 9 m
g = 9.81 m/s²
We can use the equation
Wfric = - (μ*N)*d ⇒ Wfric = - μ*m*g*d = - 0.9*m*9.81*9
⇒ Wfric = - 79.461*m
Part D.
Given
ve = 12 m/s
vn = ?
We can use the formula after the collision
vf² = 0 = vs² - 2*a*d ⇒ vs = √(2*a*d)
where a can be obtained as follows
Ffric = 2*m*a ⇒ a = Ffric/(2*m) = μ*(2*m*g)/(2*m)
⇒ a = μ*g = 0.9*9.81 m/s² = 8.829 m/s²
then
vs = √(2*8.829 m/s²*9 m) = 12.6064 m/s
Now, we can apply the equation
vs = (ve + vn)/2 ⇒ vn = 2*vs - ve
⇒ vn = 2*(12.6064 m/s) - (12 m/s)
⇒ vn = 13.2128 m/s
Then, the driver N has reported his speed correctly.
Answer:
a) [tex]v^2 = (\frac{v_n}{2})^2 + (\frac{v_e}{2})^2[/tex]
b) [tex]K = {m}((\frac{v_n}{2})^2 + (\frac{v_e}{2})^2)[/tex]
c) [tex]W_f_r_i_c = -\mu 2mgd[/tex]
d) 22 m/s
Explanation:
a) The eastbound car is your x-component momentum, while the northbound car is your y-component momentum.
[tex]p_x = mv_e_i[/tex]
[tex]p_y = mv_n_i[/tex]
Now, use the Momentum Conservation Principle. Since both cars have equal mass, both will now be referred to as solely m.
[tex]2mv_e_f = mv_e_i[/tex]
Solve for [tex]v_e_f[/tex]
[tex]v_e_f = \frac{v_e_i}{2}[/tex]
Do the same steps to solve for [tex]v_n_f[/tex]
[tex]2mv_n_f = mv_n_i[/tex]
[tex]v_n_f = \frac{v_n_i}{2}[/tex]
Then, use the Pythagorean Theorem to find [tex]v_f[/tex] & finally get [tex]v^2[/tex]
[tex]v_f = \sqrt{(\frac{v_e}{2})^2 + (\frac{v_n}{2})^2[/tex]
[tex]v^2 = (\frac{v_e}{2})^2 + (\frac{v_n}{2})^2[/tex]
b) Fairly straightforward here, just use the kinetic energy formula. Since both cars have collided, you get 2m & that cancels with the 2 on the bottom.
[tex]KE = \frac{mv^2}{2}[/tex]
[tex]K = m((\frac{v_e}{2})^2 + (\frac{v_n}{2})^2)[/tex]
c) [tex]W = Fd[/tex]
[tex]F_f_r_i_c = -\mu 2mg[/tex]
[tex]W_f_r_i_c = -\mu 2mgd[/tex]'
d) K = W (Work is no longer negative here)
[tex]m((\frac{v_e}{2})^2 + (\frac{v_n}{2})^2) = \mu 2mgd[/tex]
Cancel the m's & distribute the exponents
[tex]\frac{v_e^2}{4} + \frac{v_n^2}{4} = \mu 2gd[/tex]
Multiply both sides by 4 & plug in your correlating values
[tex]12^2 + v_n^2 = (0.9)(8)(9.81)(9)[/tex]
Then, simply solve for [tex]v_n[/tex]
[tex]v_n = \sqrt{(0.9*8*9.81*9)-144} =>[/tex] 22 m/s