ABSTRACT

The objective of this experiment was to find how the vertical distance the ball drops is related to the horizontal distance the ball travels when launched horizontally.Horizontal and vertical distances were measured. Time of flight and initial velocity was calculated by different methods. Different reasons for the errors were noticed.

Experiment 6 (Conservation of Momentum in two dimensions):

The objective of this experiment was to note and show that momentum and energy are conserved in an elastic collision and an inelastic collision. Two balls were collided. First the conservation of momentum and energy for elastic collision was observed and then for the inelastic collision. Several reasons for errors were noticed.

Equipment:

The equipment needed in this experiment was:

– Mini Launcher and steel ball

– Measuring tape or meter stick

– Carbon paper

– White paper

– Movable vertical target board (Must reach from floor to muzzle)

– Graph paper.

ABSTRACTThe objective of this experiment was to find how the vertical distance the ball drops is related to the horizontal distance the ball travels when launched horizontally.Horizontal and vertical distances were measured. Time of flight and initial velocity was calculated by different methods. Different reasons for the errors were noticed.

Formulas:Time of flight of Ball:

1

The vertical distance was calculated by:

2

(from eq. 1 and 2)

So,y is proportional to x2

Theory:A projecile follows an elliptical path which is clear from eq A. now we want to check this relation by experiment. From eq A y is directly related to x2 which can be used to check the relationship of eq A. in this experiment we used a vertical target to take different values of y with respect to x.

Procedure:In this experiment we clamped the launcher and placed the whole apparatus on the floor.Then we adjusted the angle of launcher to zero degree to launch the ball horizontally. Then we fired a ball on the target board keeping it very

close to the launcher for determining its orignal height (yo). Then we fired a test shot to determine the position of the target and placed a white and carbon paper on it at that place. The board was palced such that the ball was hitting bottom of the paper attached to it. After this we fired the ball and measured the horizontal and vertical distance of the ball. After this we decreased the distance of board from launcher and repeated the experiment. We kept on decreasing distance of board till it was only 10 cm below muzzle.

DATA ANALYSISSerial no Horizontal

distance (x) mVertical distance(y-yo) m

X2

1 1.454 -0.236 2.1142 1.354 -0.172 1.8333 1.254 -0.152 1.5724 1.154 -0.122 1.3315 1.054 -0.105 1.1106 0.954 -0.076 0.9107 0.854 -0.059 0.7298 0.754 -0.041 0.5689 0.654 -0.03 0.427

10 0.554 -0.013 0.30611 0.454 -0.007 0.20612 0.354 -0.006 0.12513 0.254 -0.002 0.064

Graph of y w.r.t x 2

Equipment:

The equipment needed in this experiment was:

– Mini Launcher.

- 2 steel balls and Collision Attachment.

– Plumb bob.

– Meter stick .

– Protractor.

– Butcher paper.

– Tape to make collision inelastic.

– Stand to hold ball.

– Carbon paper.

Theory:According to law of conservation of momentum;

Initial momentum = final momentum

Initially we would fire a single ball whose momentum would be m1v1 and finally we would strike the ball with another ball and their momentum would be ball 1 (m 1v3) and for ball 2 (m2v2).

So, m1v1 = m1v3 + m2v2

we took both the balls of same mass so the eq would reducev1 = v3 + v2

as the height of both the balls are same so they would take same time for landing as

t =√ 2hgso the eq would reduce to X1 = X2 + X3

Same the law of conservation of energy eq would reduce asX12= X2

2 + X32.

These two eq. would be tested for one-dimensional and 2 dimensional case.

As energy is a scalar quantity so it would not matter whether the motion of the balls is along same direction or different. But for momentum it matters and we have to take its resolved components separately for 2 dimensions.

At the end in inelastic case only law of conservation of energy holds which would be checked by the same eq.

Formulas:1 dimensional

The formulas used in experiment were as follows: For conservation of momentum we check the following eq.

X1 = X2 + X3

For checking conservation of kinetic energy we would check the relation

X1 2= X2

2 + X32

2 dimensional

For Momentum, For this purpose the motion or the distances would be divided in their components and would be checked seperatly.

The eq are for x-axis X1 = X2 + X3

for y-axis Y1 = Y2 + Y3

For kinetic energy, As energy is a scalar quantity we would take the absolute values of the distances as in unidimensional case.

Procedure:1. The mini launcher was kept on the floor.2. Angle of the launcher would be kept at zero degree for whole experiment.3. A ball was launched to find the initial distance X1.4. Now a second ball of the same mass was placed in front of the launcher tube with

the help of a T. and the ball 1 was launched again which would hit the ball 2 and the final distances was taken.

5. For 1 dimensional we placed the ball 2 in front of the middle of the launching tube so the ball 1 hitted ball 2 at its center and both the balls went straight.

6. For 2 dimensional case we placed the ball such that ball 1 stroke a side of ball 2 and both the balls were diverged. We took both the distances X1 , X2 and also the normal distance of final points to the middle point for calculating the angle.

7. At the end we tapped the ball 2 ont the T and took a reading for inelastic case.Note: while taking all the readings first we fired the balls to check their average landing position. Then we placed a white paper with a carbon paper tapped on it at that place and took 4 readings. The average of these readings was considered to be true.

DATA ANALYSIS

A. One dimension momentum test: i. Initial distance: X 1:

Distance from launcher and paper edge = 184.8 = 1.848

Serial no Δx1 (cm)1 10.12 13.43 14.24 14.9Average (cm) 13.15Average (meters) 0.132X1 = x1 + Δx1 1.98X1

2 3.9204

ii. Straight ball shooting far point: X 2:

Distance from paper and mini launcher = x2 = 187.8 cm = 1.878 m

Serial no Δx2 (distance of point to paper edge) cm

1 2.252 0.93 2.54 1.75 1.15Average (cm) 1.838Average (meters) 0.0184Total (X2) = x2 + average Δx2 1.896X2

2 3.595

iii. Straight ball shooting close point X 3: Distance from launcher to the paper edge = x3 = -4 cm = -0.04 m

Serial no Δx3 cm1 11.552 10.53 8.94 8.05 5.16 8.357 1.1Average (Δx3) cm 7.64Average (meters) 0.0764Total distance(X3) 0.0364X3

2 0.00132Error in momentum:X2 + X3 = 1.896 + 0.0364 = 1.9324

Error% = 1.98– 1.9324

1.98*100 = 2.4%

Error in kinetic energy:X22 + X3

2 = 3.595 + 0.00132 = 3.59632

Error% = 3.9204−3.59632

3.9204*100 = 8.27%

B. 2 dimensional final momentum: i. Left side ball:D 2:

Distance of launcher and paper edge = x2 = 96 cm = 0.96 m

Serial no Δd cm1 152.52 1283 11.654 9.9

5 3.956 1.3average 51.22Average (meters) 0.512Total distance (meters) D2 1.472D2

2 2.167ii. Right side ball:D 3:

Distance of launcher and paper edge = x3 = 129 cm = 1.29 m

Serial no. Δd1 11.52 123 8.44 2.755 2.46 0.657 0.5Average 5.46Average (meters) 0.0546Total distance (meters) D3 1.345D3

2 1.809iii. Angle:

Θ3 = tan-1 (73103.2

) = 43.12

Θ2 = tan-1 (86.6138.4

) = 32.0

For final momentum along x-axis:D2. CosѲ2 + D3. CosѲ3 = 1.472 * Cos(32) + 1.345 * Cos(43.12)= 2.338Error in momentum along x-axis

Error% = 1.98−2.3381.98

*100 = -18.08%For final momentum along y-axisD2. SinѲ2 + D3. SinѲ3 = 1.472 * Sin(32) - 1.345 * Sin(43.12)= 0.134which is the error in y direction because Y initial is zero.

For final kinetic energy:D2

2 + D32 = 1.4722 + 1.3452= 3.976

Error% = 3.9204−3.9763.9204

*100 = 1.43%

C.Inelastic case

x=0.15mx́=0.227my1=0.102m

y2=0.136mθ1=42.8θ2=36.8

x=0.320mx1=x cosθ1=(0.15 ) (0.73 )=0.110m

x2= x́ cosθ2=(0.227 ) (0.82 )=0.181mx1+ x2=0.292m

Initial X momentum

0.320 Final X momentum

0.292 % difference 8.8%

Y momentum ball 1

0.102 Y momentum ball 2

0.136 % difference 25%

Initial KE 0.1024 Final KE 0.0737 % difference 28%

CONCLUSION

In first experiment we learnt how vertical distance is related to horizontal distance in a projectile motion. We calculated error and got to know about the factors which affected our experiment.

In 2nd experiment we learnt different cases like elastic and inelastic collision, and relationship between kinetic energy and momentum in elastic and inelastic collision.

We observed that momentum (both in x and y) and energy are conserved in an elastic collision.

We also observed that only momentum is conserved in an inelastic collision. Energy is not conserved in this case. That is why we got large error in energy for inelastic collision.

In this experiment a limitation which we observed was that the tee was inclined at a particular angle so that the ball went further than the original distance when hit.

Causes of error1. Resistance provided by the T.2. When the ball was placed on the T it was a little higher than the

targeting ball. Therefore ball 2 was got launched on an angle greater than zero and travelled more distance (range).