WEBVTT
00:00:00.230 --> 00:00:06.830
A tennis racquet hits a tennis ball that has a mass of 60.5 grams and applies a constant 75-newton force to it.
00:00:07.340 --> 00:00:11.210
The tennis ball changes its velocity by 30 meters per second during the collision.
00:00:11.700 --> 00:00:13.780
For how many seconds does the collision last?
00:00:14.590 --> 00:00:17.530
Okay, so to answer this question, let’s start out by drawing a diagram.
00:00:17.970 --> 00:00:21.100
So here’s a rather badly drawn side-on view of a tennis racquet.
00:00:21.600 --> 00:00:23.790
And here’s our tennis racquet hitting a tennis ball.
00:00:24.330 --> 00:00:26.510
Now let’s say that this tennis ball has a mass 𝑚.
00:00:27.120 --> 00:00:31.410
We’ve been told in the question that 𝑚 is equal to 60.5 grams.
00:00:32.060 --> 00:00:36.020
As well as this, let’s say that the tennis racquet exerts a force 𝐹 on the ball.
00:00:36.610 --> 00:00:40.760
Once again, we’ve been told in the question that the value of 𝐹 is 75 newtons.
00:00:41.350 --> 00:00:46.840
Let us also say that the change in velocity of the ball, which is caused by the collision with the racquet, is Δ𝑣.
00:00:47.210 --> 00:00:51.310
And we know once again from the question that Δ𝑣 is 30 meters per second.
00:00:52.000 --> 00:00:59.130
What we’ve been asked to do is to find out the amount of time or for how many seconds the collision between the tennis ball and the tennis racquet lasts.
00:00:59.470 --> 00:01:06.060
In other words, let’s say that Δ𝑡, the time interval for which the collision between the racquet and the ball lasts, is as of yet unknown to us.
00:01:06.720 --> 00:01:08.960
Now we need to go about working out the value of Δ𝑡.
00:01:09.620 --> 00:01:18.970
To do this, we can recall that an impulse on an object is defined as the force applied on an object multiplied by the time interval, Δ𝑡, for which that force acts on the object.
00:01:19.550 --> 00:01:21.610
So in this case, our object is the tennis ball.
00:01:22.110 --> 00:01:31.130
And a tennis racquet is applying a force 𝐹 onto the tennis ball for a time interval Δ𝑡, which is the amount of time for which the collision between the ball and the racquet occurs.
00:01:31.870 --> 00:01:37.270
Now the other thing that we can also recall is that the impulse is equal to the change in momentum of the object, Δ𝑝.
00:01:37.950 --> 00:01:40.450
At this point, however, we don’t know what Δ𝑝 is.
00:01:40.720 --> 00:01:47.160
But we can recall that the momentum of an object is given by the mass of the object multiplied by the velocity of the object.
00:01:47.670 --> 00:01:54.540
So we can recall that the change in momentum, Δ𝑝, of that object is given as the change in mass times velocity of the object.
00:01:55.090 --> 00:01:57.420
However, in this case, our object is the tennis ball.
00:01:57.510 --> 00:02:00.170
And the mass of that tennis ball is not changing during the collision.
00:02:00.730 --> 00:02:04.980
So we can say that the change in momentum of the ball is equal to 𝑚 multiplied by Δ𝑣.
00:02:05.610 --> 00:02:08.000
This is because only the velocity changes at all.
00:02:08.520 --> 00:02:12.450
And remember, this expression is only true for when the mass of the object stays constant.
00:02:13.110 --> 00:02:15.380
Now at this point, we can work out the value of Δ𝑝.
00:02:15.850 --> 00:02:19.320
Δ𝑝 is equal to the mass of the ball, which is 60.5 grams.
00:02:19.730 --> 00:02:20.980
But let’s not use grams.
00:02:20.980 --> 00:02:23.260
Let’s convert it to the standard unit of kilograms.
00:02:23.880 --> 00:02:27.570
To do this, we recall that one gram is equal to one thousandth of a kilogram.
00:02:28.030 --> 00:02:35.340
So multiplying both sides of the equation by 60.5, we find that 60.5 grams is equal to 0.0605 kilograms.
00:02:35.840 --> 00:02:42.160
And hence, we can change the mass to 0.0605 kilograms, which means we can come back to finding out our change in momentum.
00:02:42.710 --> 00:02:54.170
So once again, the change in momentum is equal to the mass of the tennis ball, which we now know to be 0.0605 kilograms, multiplied by the change in velocity of the object, which is 30 meters per second.
00:02:54.220 --> 00:02:56.290
And this is already in its standard unit.
00:02:56.940 --> 00:03:00.330
Now we had kilograms for the mass and meters per second for the velocity.
00:03:00.570 --> 00:03:16.390
So our change in momentum is going to be in a unit of kilogram meters per second, which once again is the standard unit of momentum or change in momentum, at which point when we evaluate this side of the equation, we find that the change in momentum is equal to 1.815 kilograms meters per second.
00:03:16.850 --> 00:03:19.520
At this point, we have a value for the change in momentum, Δ𝑝.
00:03:19.810 --> 00:03:21.950
And we already knew what 𝐹, the force, was.
00:03:22.520 --> 00:03:25.670
So we can work out the value of Δ𝑡 by rearranging this equation.
00:03:26.330 --> 00:03:32.090
We first write down in our table of important information that the change in momentum is 1.815 kilograms meters per second.
00:03:32.370 --> 00:03:36.880
And then we rearrange to find Δ𝑡 in this equation by dividing both sides of the equation by 𝐹.
00:03:37.300 --> 00:03:39.610
Doing this leaves us with Δ𝑡 on the left-hand side.
00:03:39.610 --> 00:03:44.710
And on the right, we have Δ𝑝 divided by 𝐹, at which point we just need to plug in values.
00:03:45.130 --> 00:03:51.290
So we have the change in momentum, 1.815 kilograms meters per second, divided by the force, 75 newtons.
00:03:51.810 --> 00:04:01.720
And once again, because we’re working at standard units on the right-hand side, we’ve got standard units for both change in momentum and force, we’re going to get the time interval in its standard unit of seconds.
00:04:02.390 --> 00:04:09.060
So evaluating the right-hand side of the equation, we find that the time interval is equal to 0.0242 seconds.
00:04:09.620 --> 00:04:11.730
However, this is not our final answer.
00:04:12.300 --> 00:04:15.640
This is because we need to round our answer to the correct number of significant figures.
00:04:16.290 --> 00:04:25.920
Now in the question, we’ve been given one value to three significant figures and two of them to two significant figures, because both 75 and 30 are to two significant figures.
00:04:26.410 --> 00:04:29.110
Two significant figures, that’s what we need to round our answer to.
00:04:29.660 --> 00:04:32.780
So here’s the first significant figure and here’s the second.
00:04:33.280 --> 00:04:38.720
Now it’s the significant figure afterwards the third significant figure that will tell us what happens to the second significant figure.
00:04:39.380 --> 00:04:41.660
In this case, the third significant figure is a two.
00:04:42.140 --> 00:04:43.510
Two is less than five.
00:04:44.060 --> 00:04:46.340
Therefore, the second significant figure will stay the same.
00:04:46.340 --> 00:04:47.670
It’s not gonna round up.
00:04:48.300 --> 00:04:52.720
Hence, to two significant figures, our time interval is 0.024 seconds.
00:04:53.200 --> 00:04:54.630
And so we have our final answer.
00:04:55.030 --> 00:04:58.820
The collision lasts for 0.024 seconds to two significant figures.