Question

cally up from a moving cart. The position of the balls to raph below shows the position of the ball at equal time inter- A ball is fired vertically up from a mo the left of the graph below showst vals, as would be seen by the person on the cart. a. For this person the motion is sym- metric in time so positions on the way down are the same as on the way up, so only one ball is drawn for both up and down positions. Draw the ball at each time to tot, as seen by the lab observer. The positions at the times to and t, are shown. VERTICAL DIRECTION (POSITIVE y-AXIS) Explain the basis for your predic- tion. HORIZONTAL DIRECTION (POSITIVE X-AXIS) Figure 5-8 b. In a second experiment the cart is given an identical push as above, so has the same velocity, but the launcher has a stronger spring thereby increasing the muzzle velocity of the ball. How does this affect the motion and path of the ball? Increase/decrease/stay the same? i. The horizontal velocity? ii. The horizontal distance traveled (range)? iii. The vertical height attained iv. The time in the air? In a third experiment the cart is pushed harder instead, so it has a greater velocity, while the muzzle velocity is the same as in the first experiment. In comparison with the first experiment, do the following increase/decrease/stay the same? i The time in the air ii. The vertical velocity of the ball? iii. The horizontal distance traveled (range)? iv. The vertical height attained?

Answer 1

PROBLEM 1:

__________________________________________________________________________

PART 1:

The solution diagram is

The fact that there is no acceleration in the horizontal direction makes its velocity constant. So, it moves equal distance on the x-axis in equal time intervals.

__________________________________________________________________________

PART B:

(i) The horizontal velocity stays the same.

(ii) The horizontal distance traveled (range) increases.

(iii) The vertical height attained increases.

(iv) The time in the air increases.

__________________________________________________________________________

PART C:

(i) The time in the air stays the same.

(ii) The vertical velocity of the ball stays the same.

(iii) The horizontal distance traveled (range) increases.

(iv) The vertical height attained stays the same.

__________________________________________________________________________

NOTE: Please reupload other questions saperately to get their solutions. One question at a time is preffered.

Answer 2

Ans to Q no. 1:

a.

In this case, there is no acceleration in the horizontal direction. So, the component of velocity in the horizontal direction would not change. But, there is acceleration due to gravity in the vertical direction. So, the vertical velocity will gradually decrease, become zero and increase again in the negative Y direction.

b.

In the second case, the horizontal velocity of the cart is same as above. So, the horizontal velocity profile would not change in this case.

So,

i. The horizontal velocity stay the same.

Since, the initial vertical velocity is more in this case, but the acceleration in the vertical direction is same as above (the acceleration due to gravity), the ball would travel a higher distance in the vertical direction.

So,

ii. The horizontal distance traveled (range) increases.

iii. The vertical height attained increases.

iv. The time in the air increases.

c.

In this case, the cart is pushed harder. So, the horizontal velocity of the cart increases. But the initial vertical velocity of the ball is same as the first case.

So,

i. The time in the air remains the same.

ii. The vertical velocity of the ball increases.

iii. The horizontal distance traveled (range) increases.

iv. The vertical height attained remains the same.