A particle is projected from the origin $(0, 0)$ $(0,0)$ in the $x y$ $xy$ plane at angle of $37^{o}$ $37^o$ with the x axis with a speed of $10$ $10$ $m s^{- 1}$ $ms^-1$ , its acceleration being $(- 5 ˆ i + 10 ˆ j)$ $(-5 hati + 10 hatj)$ $m s^{- 2}$ $ms^-2$ . What is the speed of the particle when the x-coordinate is again zero?

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A particle is projected from the origin $(0, 0)$ $(0,0)$ in the $x y$ $xy$ plane at angle of $37^{o}$ $37^o$ with the x axis with a speed of $10$ $10$ $m s^{- 1}$ $ms^-1$ , its acceleration being $(- 5 ˆ i + 10 ˆ j)$ $(-5 hati + 10 hatj)$ $m s^{- 2}$ $ms^-2$ . What is the speed of the particle when the x-coordinate is again zero?

Answer given in the book is $22$ $22$ $m s^{- 1}$ $ms^-1$ .

Please explain with steps.

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Answer 1 · 2017-06-11T22:51:38

$39 \frac{m}{s}$ $39"m"/"s"$

Explanation:

We can split this problem up into $x$ $x$ - and $y$ $y$ -components.

We can find the components of the initial velocity using the equation

$v_{0 x} = v_{0} cos α$ $v_(0x) = v_0cosalpha$

for $x$ $x$ , and

$v_{0 y} = v_{0} sin α$ $v_(0y) = v_0sinalpha$

for $y$ $y$ . The initial velocity components are thus

$v_{0 x} = (10 \frac{m}{s}) cos (37^{o}) = 7.99 \frac{m}{s}$ $v_(0x) = (10"m"/"s")cos(37^"o") = 7.99"m"/"s"$

$v_{0 y} = (10 \frac{m}{s}) sin (37^{o}) = 6.02 \frac{m}{s}$ $v_(0y) = (10"m"/"s")sin(37^"o") = 6.02"m"/"s"$

We're given that the equations for the components of acceleration are

$a_{x} = - 5 \frac{m}{s^{2}}$ $a_x = -5"m"/("s"^2)$

$a_{y} = 10 \frac{m}{s^{2}}$ $a_y = 10"m"/("s"^2)$

Using the initial velocity components, and integrating these acceleration equations, we have for the velocity equations

$v_{x} = 7.99 \frac{m}{s} - (5 \frac{m}{s^{2}}) t$ $v_x = 7.99"m"/"s" - (5"m"/("s"^2))t$

$v_{y} = 6.02 \frac{m}{s} + (10 \frac{m}{s^{2}}) t$ $v_y = 6.02"m"/"s" + (10"m"/("s"^2))t$

Integrating these equations, the position equations are

$x = (7.99 \frac{m}{s}) t - \frac{1}{2} (5 \frac{m}{s^{2}}) t^{2}$ $x = (7.99"m"/"s")t - 1/2(5"m"/("s"^2))t^2$

$y = (6.02 \frac{m}{s}) t + \frac{1}{2} (10 \frac{m}{s^{2}}) t^{2}$ $y = (6.02"m"/"s")t + 1/2(10"m"/("s"^2))t^2$

We must find the time when $x = 0$ $x = 0$ , so

$x = (7.99 \frac{m}{s}) t - \frac{1}{2} (5 \frac{m}{s^{2}}) t^{2}$ $x = (7.99"m"/"s")t - 1/2(5"m"/("s"^2))t^2$

$t = 3.19$ $t = 3.19$ $s$ $"s"$

Using this time, we can find the components of the velocity by

$v_{x} = 7.99 \frac{m}{s} - (5 \frac{m}{s^{2}}) (3.19 s) = - 7.99 \frac{m}{s}$ $v_x = 7.99"m"/"s" - (5"m"/("s"^2))(3.19"s") = -7.99"m"/"s"$

$v_{y} = 6.02 \frac{m}{s} + (10 \frac{m}{s^{2}}) (3.19 s) = 38.0 \frac{m}{s}$ $v_y = 6.02"m"/"s" + (10"m"/("s"^2))(3.19"s") = 38.0"m"/"s"$

The magnitude of the velocity (the speed) is thus

$v = \sqrt{{(- 7.99 \frac{m}{s})}^{2} + {(38.0 \frac{m}{s})}^{2}} = 39 \frac{m}{s}$ $v = sqrt((-7.99"m"/"s")^2 + (38.0"m"/"s")^2) = color(red)(39"m"/"s"$

rounded to $2$ $2$ significant figures, the amount given in the problem.

Correct me if I made a mistake, but I think this answer should be correct.

Answer 2 · 2017-06-11T23:04:06

$v = \sqrt{v_{ˆ i}^{2} + v_{ˆ j}^{2}} = \sqrt{{(- 8)}^{2} + 38^{2}} = 39$ $v = sqrt(v_hati^2+v_hatj^2)=sqrt((-8)^2+38^2)=39$ $m s^{- 1}$ $ms^-1$

See the detailed explanation and working below.

Explanation:

(the units given are a bit odd, in that they mix $x$ $x$ and $y$ $y$ with $ˆ i$ $hati$ and $ˆ j$ $hatj$ , but remember that $ˆ i$ $hati$ is a vector in the $x$ $x$ direction and $ˆ j$ $hatj$ a vector in the $y$ $y$ direction and it's OK)

So our object is projected from the origin into the first quadrant - top right side of the Cartesian plane, at $37^{o}$ $37^o$ to the x-axis and $10$ $10$ $m s^{- 1}$ $ms^-1$ . It has an acceleration upward (the positive coefficient of $ˆ j$ $hatj$ ) and to the left (the negative coefficient of $ˆ i$ $hati$ , so it is going to move upward and curve back to the left, eventually crossing the y-axis (the line $x = 0$ $x=0$ ) again.

This is a different picture from our usual projectile motion problems, because the acceleration is usually provided by gravity and is downward, but it's still just a projectile motion problem, with the exception that there are accelerations in both directions, not just one. We don't know, or need to know, the mass of the particle, because gravity is not involved. We are told its acceleration.

The components of the initial velocity will be given by $u_{ˆ i} = 10 {cos 37}^{o} = 8.0$ $u_hati=10 cos37^o=8.0$ $m s^{- 1}$ $ms^-1$ and $u_{ˆ j} = 10 {sin 37}^{o} = 6.0$ $u_hatj=10sin37^o=6.0$ $m s^{- 1}$ $ms^-1$ .

We can consider each dimension independently. The position in the x direction will reach a maximum then return to $x = 0$ $x=0$ . The velocity will go to zero in the x direction at that maximum displacement, at half the time before the particle returns to $x = 0$ $x=0$ .

$v = u + a t$ $v = u + at$

$t = \frac{v_{ˆ i} - u_{ˆ i}}{a_{ˆ i}} = \frac{0 - 8}{- 5} = 1.6$ $t = (v_hati-u_hati)/a_hati = (0-8)/-5 =1.6$ $s$ $s$

So the total time will be $3.2$ $3.2$ $s$ $s$ .

The final velocity in the x direction, $v_{ˆ i}$ $v_hati$ , will be:

$v_{ˆ i} = u_{ˆ i} + a_{ˆ i} t = 8 + (- 5) 3.2 = - 8$ $v_hati=u_hati+a_hatit = 8+(-5)3.2=-8$ $m s^{- 1}$ $ms^-1$

The final velocity in the y direction, $v_{ˆ j}$ $v_hatj$ , will be:

$v_{ˆ j} = u_{ˆ j} + a_{ˆ j} t = 6 + (10) 3.2 = 38$ $v_hatj=u_hatj+a_hatjt = 6+(10)3.2=38$ $m s^{- 1}$ $ms^-1$

The magnitude of the final velocity will be given by:

$v = \sqrt{v_{ˆ i}^{2} + v_{ˆ j}^{2}} = \sqrt{{(- 8)}^{2} + 38^{2}} = 39$ $v = sqrt(v_hati^2+v_hatj^2)=sqrt((-8)^2+38^2)=39$ $m s^{- 1}$ $ms^-1$

This is not the answer given in the question, but other answerers get the same, so I assume the textbook is incorrect.

Answer 3 · 2017-06-12T00:29:49

I made my own attempt at this and got $39 m/s$ $"39 m/s"$ without gravity in the original coordinates.

I think the book forgot the $\frac{1}{2}$ $1/2$ in front of the acceleration in using the 2D kinematics equations.

DISCLAIMER: LONG ANSWER!

Notice how if you invert the coordinate axes, i.e. reflect over $y = x$ $y = x$ , you get a normal scenario, except with a nonzero horizontal acceleration (after the inversion).

GIVEN DATA IN NEW COORDINATES

We will do that, and obtain:

${\to v}_{i} = 10 m/s$ $vecv_i = "10 m/s"$
${\to a}_{x} = {10 m/s}^{2}$ $veca_x = "10 m/s"^2$
${\to a}_{y} = - {5 m/s}^{2}$ $veca_y = -"5 m/s"^2$
${\to x}_{i} = 0 m$ $vecx_i = "0 m"$
${\to y}_{i} = 0 m$ $vecy_i = "0 m"$
$θ = 90^{\circ} - 37^{\circ} = 53^{\circ}$ $theta = 90^@ - 37^@ = 53^@$ with respect to (w.r.t) the new $x$ $x$ axis

Instead of solving for when $x = 0$ $x = 0$ , we now solve for when we cross the $x$ $x$ axis again.

DERIVATION OF KINEMATICS EQUATIONS

Integrate the acceleration w.r.t. time:

${\to v}_{f x} = \int {\to a}_{x} d t = {\to a}_{x} t + {\to v}_{i x}$ $vecv_(fx) = int veca_x dt = veca_(x)t + vecv_(ix)$ $(1)$ $" "" "bb((1))$

${\to v}_{f y} = \int {\to a}_{y} d t = {\to a}_{y} t + {\to v}_{i y}$ $vecv_(fy) = int veca_y dt = veca_(y)t + vecv_(iy)$ $(2)$ $" "" "bb((2))$

Integrate one more time w.r.t time:

${\to x}_{f} = \int {\to a}_{x} t + {\to v}_{i x} d t = \frac{1}{2} {\to a}_{x} t^{2} + {\to v}_{i x} t$ $vecx_f = int veca_(x)t + vecv_(ix) dt = 1/2veca_(x)t^2 + vecv_(ix)t$ $(3)$ $" "" "bb((3))$

${\to y}_{f} = \int {\to a}_{y} t + {\to v}_{i y} d t = \frac{1}{2} {\to a}_{y} t^{2} + {\to v}_{i y} t$ $vecy_f = int veca_(y)t + vecv_(iy) dt = 1/2veca_(y)t^2 + vecv_(iy)t$ $(4)$ $" "" "bb((4))$

since $({\to x}_{i}, {\to y}_{i}) = (0, 0)$ $(vecx_i, vecy_i) = (0,0)$ .

SOLVING FOR THE TIME OF FLIGHT

Now, using $(4)$ $(4)$ , solve for the time when we cross the $x$ $x$ axis, i.e. when ${\to y}_{f} = 0$ $vecy_f = 0$ . Cross out the trivial $t = 0$ $t = 0$ to get:

$\frac{1}{2} {\to a}_{y} t + {\to v}_{i y} = 0$ $1/2veca_(y)t + vecv_(iy) = 0$

$\Rightarrow t = - \frac{2 {\to v}_{i y}}{{\to a}_{y}}$ $=> t = -(2vecv_(iy))/(veca_y)$

$= -(2cdot"10 m/s" cdot sin(53^@))/(-"5 m/s"^2)$

$=$ $"3.195 s"$

VELOCITY COMPONENTS

This means from $(1)$ and $(2)$ , we can solve for the final velocity components:

$vecv_(fx) = ("10 m/s"^2)("3.195 s") + "10 m/s" cdot cos(53^@)$

$=$ $"37.96 m/s"$

$vecv_(fy) = (-"5 m/s"^2)("3.195 s") + "10 m/s" cdot sin(53^@)$

$=$ $-"7.986 m/s"$

FINAL VELOCITY

The final velocity I get is:

$color(blue)(vecv_f) = sqrt((37.96)^2 + (-7.986)^2)$

$=$ $color(blue)("39 m/s") ne 22$

ERROR IN THE BOOK!

Interestingly enough, if we forgot the $1/2$ in front of the kinematics equations we derived...

$=> t = -(color(red)(1)vecv_(iy))/(veca_y)$

$= -(color(red)(1)cdot"10 m/s" cdot sin(53^@))/(-"5 m/s"^2)$

$=$ $"1.597 s"$

And we would have gotten:

$vecv_(fx) = ("10 m/s"^2)("1.597 s") + "10 m/s" cdot cos(53^@)$

$=$ $"21.991 m/s"$

$vecv_(fy) = (-"5 m/s"^2)("1.597 s") + "10 m/s" cdot sin(53^@)$

$=$ $-1.38 xx 10^(-31)$ $"m/s" ~~ 0$

The final velocity I get would then be:

$color(blue)(vecv_f) ~~ vecv_(fx)$

$=$ $color(blue)("22 m/s")$

So yes, I think the book has an error...

Answer 4 · 2017-06-12T01:01:51

The $x$ and $y$ direction being orthogonal to each other can be treated independently. From the expression for acceleration we see that it is a constant acceleration in both axes.

Movement along $x$ axis

$u_x=10cos 37$
$a_x=-5ms^-2$
Using the kinematic expression
$s=s_0+ut+1/2at^2$
we get
$x=u_xt+1/2a_xt^2$
$x=(10cos 37)t-5/2t^2$
Using the given condition we get
$0=(10cos 37)t-5/2t^2$
Ignoring root $t=0$ being initial condition we get
$t_(x=0)=2/5xx(10cos 37)approx 3.19s$

Now using the kinematic expression
$v_x=u_x+at$ .....(1)
we get for above value of $t$
$v_x=10cos 37+(-5)xx3.19approx-8ms^-1$

Movement along $y$ -axis

$u_y = 10sin 37$
$a_y = 10ms^-2$

using (1) we get at the requited $t$
$v_y=10sin 37+10xx3.19$
$=>v_y=6.02+16approx 38ms^-1$

Speed
$|v|_(x=0)=sqrt(v_x^2+v_y^2)$
$|v|_(x=0)=sqrt(38^2+(-8)^2)$

$=>|v|_(x=0)=38.8ms^-1$

Hope this helps.

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