physics question

14

A student predicts that a steel ball dropped from rest from a height of 2 m will hit the floor
after 0.63 seconds. In their calculation they used an approximate value for the acceleration
due to gravity of g = 10 m/s2
.
The accepted value for the acceleration due to gravity is g = 9.8 m/s2
.
This means their calculated time will be:
A. unaffected
B. too long by about 1%
C. too short by about 1%
D. too long by about 0.2 seconds
E. too short by about 0.2 seconds

Reply 1

GgbroTG

10

Consider the equation of motion for an object under constant acceleration:

$[br]x = ut + 1/2at^2[br]$

You should recognise this as one of the "suvat" equations (with x instead of s by convention).

If the ball is dropped, then the initial velocity ( $u$ ) is 0.

Try rearranging the formula, see if you can solve for $t$ , then you should be able to find out how $t$ depends on $a$ .

Reply 2

Stonebridge

13

Original post

by Ashirs

A student predicts that a steel ball dropped from rest from a height of 2 m will hit the floor
after 0.63 seconds. In their calculation they used an approximate value for the acceleration
due to gravity of g = 10 m/s2
.
The accepted value for the acceleration due to gravity is g = 9.8 m/s2
.
This means their calculated time will be:
A. unaffected
B. too long by about 1%
C. too short by about 1%
D. too long by about 0.2 seconds
E. too short by about 0.2 seconds

Write down the formula they would use to calculate this time.
It will be obtained from rearranging the equation exactly as the other poster has said.
I suggest you use the % error in a to find the % error in the value of time calculated.
Using a value of 10 instead of 9.8 introduces a % error of 2% in the value of g (say if you don't understand this part or haven't done % errors)
The next step requires you to know what happens to this % error when you have to find t, using that value of a
You also need to think whether this will make t bigger or smaller.

Reply 3

Ashirs

OP

14

ok thank you!
i used the equation distnace= root (2*s)/a
is that right?

Reply 4

GgbroTG

10

I assume you mean time, rather than distance, but yes, that's correct.

With that formula in hand, you can either propagate the percentage error as suggested by @Stonebridge (which is good practice especially if you plan to study physics at uni, for example), or substitute the two given values of $g$ (9.8 and 10) for $a$ and compare the two answers.

Intuitively, we should expect that since $t$ is proportional to $\frac{1}{\sqrt{a}}$ , a larger acceleration should result in a shorter fall time.

Reply 5

Ashirs

OP

14

yes sorry i meant time
thank you very much for your help!!!

Reply 6

kunli

1

Original post

by Ashirs

A student predicts that a steel ball dropped from rest from a height of 2 m will hit the floor
after 0.63 seconds. In their calculation they used an approximate value for the acceleration
due to gravity of g = 10 m/s2
.
The accepted value for the acceleration due to gravity is g = 9.8 m/s2
.
This means their calculated time will be:
A. unaffected
B. too long by about 1%
C. too short by about 1%
D. too long by about 0.2 seconds
E. too short by about 0.2 seconds

B too short

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