Question:
A block of mass of 1 kg is moving on the x axis. A force F acting on the block is shown. Velocity of the block at time $t = 2\, s$ is - $3 \,m/s$ What is the speed of the block at time t = 4s ?
A block of mass of 1 kg is moving on the x axis. A force F acting on the block is shown. Velocity of the block at time $t = 2\, s$ is - $3 \,m/s$ What is the speed of the block at time t = 4s ?
Updated On: Apr 26, 2024
- 5 m/s
- 8 m/s
- 2 m/s
- 3 m/s
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The Correct Option is D
Solution and Explanation
Given, mass of block $( m )=1 \,kg$
Velocity of block at time $(t) 2 s$ is $-3 m / s$
Impulse during $t=2\, s$ to $t=4 \,s$
From the graph,
$ I=$ Area of $\Delta\, ABC$ + Area of $\Delta \,CDE$
$=\frac{1}{2}(3-2) \times 5+\frac{1}{2}(4-3)(-5) $
$=\frac{1}{2} \times 1 \times 5+\frac{1}{2} \times(1)(-5)$
$=\frac{5}{2}-\frac{5}{2}=0$
We know that,
Impulse = Change in momentum
$ 0 =m v_{f}-m v_{i} $
$ \Rightarrow \, m v_{f}-m V_{i} =0$
$ \Rightarrow \,m V_{f} =m v_{i} $
$ V_{f} =v_{i}=-3 \,m / s $
or $|3 \,m / s |$
Velocity of block at time $(t) 2 s$ is $-3 m / s$
Impulse during $t=2\, s$ to $t=4 \,s$
From the graph,
$ I=$ Area of $\Delta\, ABC$ + Area of $\Delta \,CDE$
$=\frac{1}{2}(3-2) \times 5+\frac{1}{2}(4-3)(-5) $
$=\frac{1}{2} \times 1 \times 5+\frac{1}{2} \times(1)(-5)$
$=\frac{5}{2}-\frac{5}{2}=0$
We know that,
Impulse = Change in momentum
$ 0 =m v_{f}-m v_{i} $
$ \Rightarrow \, m v_{f}-m V_{i} =0$
$ \Rightarrow \,m V_{f} =m v_{i} $
$ V_{f} =v_{i}=-3 \,m / s $
or $|3 \,m / s |$
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Concepts Used:
Speed and Velocity
The rate at which an object covers a certain distance is commonly known as speed.
The rate at which an object changes position in a certain direction is called velocity.
Difference Between Speed and Velocity:
Read More: Difference Between Speed and Velocity