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A person is observing a moving ship from the shore. Another person is on top of ship’s mast. The person in the ship drops binoculars and sees it dropping straight. The person on the shore sees the binoculars taking a curved trajectory.
Classical relativity. The same motion as viewed by two different observers. An observer on the moving ship sees the binoculars dropped from the top of its mast fall straight down. An observer on shore sees the binoculars take the curved path, moving forward with the ship. Both observers see the binoculars strike the deck at the base of the mast. The initial horizontal velocity is different relative to the two observers. (The ship is shown moving rather fast to emphasize the effect.)

Calculating relative velocity: an airline passenger drops a coin

An airline passenger drops a coin while the plane is moving at 260 m/s. What is the velocity of the coin when it strikes the floor 1.50 m below its point of release: (a) Measured relative to the plane? (b) Measured relative to the Earth?

A person standing on ground is observing an airplane. Inside the airplane a woman is sitting on seat. The airplane is moving in the right direction. The woman drops the coin which is vertically downwards for her but the person on ground sees the coin moving horizontally towards right.
The motion of a coin dropped inside an airplane as viewed by two different observers. (a) An observer in the plane sees the coin fall straight down. (b) An observer on the ground sees the coin move almost horizontally.

Strategy

Both problems can be solved with the techniques for falling objects and projectiles. In part (a), the initial velocity of the coin is zero relative to the plane, so the motion is that of a falling object (one-dimensional). In part (b), the initial velocity is 260 m/s horizontal relative to the Earth and gravity is vertical, so this motion is a projectile motion. In both parts, it is best to use a coordinate system with vertical and horizontal axes.

Solution for (a)

Using the given information, we note that the initial velocity and position are zero, and the final position is 1.50 m. The final velocity can be found using the equation:

v y 2 = v 0 y 2 2 g ( y y 0 ) . size 12{v rSub { size 8{y} rSup { size 8{2} } } =v rSub { size 8{0y} rSup { size 8{2} } } - 2g \( y - y rSub { size 8{0} } \) "."} {}

Substituting known values into the equation, we get

v y 2 = 0 2 2 ( 9 . 80 m/s 2 ) ( 1 . 50 m 0 m ) = 29 . 4 m 2 /s 2 size 12{v rSub { size 8{y} rSup { size 8{2} } } =0 rSup { size 8{2} } - 2 \( 9 "." "80"" m/s" rSup { size 8{2} } \) \( - 1 "." "50"" m" - 0" m" \) ="29" "." 4" m" rSup { size 8{2} } "/s" rSup { size 8{2} } } {}

yielding

v y = 5 . 42 m/s. size 12{v rSub { size 8{y} } = - 5 "." "42"" m/s."} {}

We know that the square root of 29.4 has two roots: 5.42 and -5.42. We choose the negative root because we know that the velocity is directed downwards, and we have defined the positive direction to be upwards. There is no initial horizontal velocity relative to the plane and no horizontal acceleration, and so the motion is straight down relative to the plane.

Solution for (b)

Because the initial vertical velocity is zero relative to the ground and vertical motion is independent of horizontal motion, the final vertical velocity for the coin relative to the ground is v y = 5.42 m/s , the same as found in part (a). In contrast to part (a), there now is a horizontal component of the velocity. However, since there is no horizontal acceleration, the initial and final horizontal velocities are the same and v x = 260 m/s size 12{"v subx =260 m/s"} {} . The x - and y -components of velocity can be combined to find the magnitude of the final velocity:

v = v x 2 + v y 2 . size 12{v= sqrt {v rSub { size 8{x} rSup { size 8{2} } } +v rSub { size 8{y} rSup { size 8{2} } } } "."} {}

Thus,

v = ( 260 m/s ) 2 + ( 5 . 42 m/s ) 2 size 12{v= sqrt { \( "260"" m/s" \) rSup { size 8{2} } + \( - 5 "." "42"" m/s" \) rSup { size 8{2} } } } {}

yielding

v = 260 . 06 m/s. size 12{v="260" "." "06"" m/s."} {}

The direction is given by:

θ = tan 1 ( v y / v x ) = tan 1 ( 5 . 42 / 260 ) size 12{θ="tan" rSup { size 8{ - 1} } \( v rSub { size 8{y} } /v rSub { size 8{x} } \) ="tan" rSup { size 8{ - 1} } \( - 5 "." "42"/"260" \) } {}

so that

θ = tan 1 ( 0 . 0208 ) = 1 . 19º . size 12{θ="tan" rSup { size 8{ - 1} } \( - 0 "." "0208" \) = - 1 "." "19"º "."} {}

Discussion

In part (a), the final velocity relative to the plane is the same as it would be if the coin were dropped from rest on the Earth and fell 1.50 m. This result fits our experience; objects in a plane fall the same way when the plane is flying horizontally as when it is at rest on the ground. This result is also true in moving cars. In part (b), an observer on the ground sees a much different motion for the coin. The plane is moving so fast horizontally to begin with that its final velocity is barely greater than the initial velocity. Once again, we see that in two dimensions, vectors do not add like ordinary numbers—the final velocity v in part (b) is not ( 260 – 5 . 42 )  m/s size 12{ \( "260 – 5" "." "42" \) " m/s"} {} ; rather, it is 260 . 06 m/s size 12{"260" "." "06 m/s"} {} . The velocity’s magnitude had to be calculated to five digits to see any difference from that of the airplane. The motions as seen by different observers (one in the plane and one on the ground) in this example are analogous to those discussed for the binoculars dropped from the mast of a moving ship, except that the velocity of the plane is much larger, so that the two observers see very different paths. (See [link] .) In addition, both observers see the coin fall 1.50 m vertically, but the one on the ground also sees it move forward 144 m (this calculation is left for the reader). Thus, one observer sees a vertical path, the other a nearly horizontal path.

Questions & Answers

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The potential in a region between x= 0 and x = 6.00 m lis V= a+ bx, where a = 10.0 V and b = -7.00 V/m. Determine (a) the potential atx=0, 3.00 m, and 6.00 m and (b) the magnitude and direction of the electric ficld at x =0, 3.00 m, and 6.00 m.
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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