Responses
impulse=acceleration×time
impulse=velocity/time
impulse=force×time
impulse=force/time
11 answers
The correct mathematical representation that identifies impulse is:
impulse = force × time
Impulse is defined as the product of the average force applied to an object and the time duration over which that force is applied. This relationship is often summarized by the impulse-momentum theorem, which states that impulse is equal to the change in momentum of the object.
In a closed system, momentum is conserved. This means that the total momentum before the collision is equal to the total momentum after the collision, regardless of how the momentum is distributed between the colliding objects.
Given:
The total initial momentum of the system is:
Total initial momentum = 24 kg·m/s + 0 kg·m/s = 24 kg·m/s.
After the collision, while the total momentum remains 24 kg·m/s, the individual momenta of the balls will change due to the collision.
Therefore, the correct statement is:
The momentum of each ball changes, and the total momentum stays the same.
To find the total momentum before the collision, we need to sum up the individual momentums of the three objects:
Given momenta:
Now, calculate the total momentum:
\[ \text{Total momentum} = 110 , \text{kg⋅m/s} + (-65 , \text{kg⋅m/s}) + (-100 , \text{kg⋅m/s}) \]
Calculating step by step:
So, the total momentum before the collision is \( -55 , \text{kg⋅m/s} \).
Since momentum is conserved in a closed system, the total momentum after the collision will also be:
−55 kg⋅m/s.
Therefore, the correct answer is:
−55 kg⋅m/s.
To find the mass of the second object, we can use the principle of conservation of momentum. The total momentum before the collision equals the total momentum after the collision.
Given:
The total momentum after the collision can be expressed as: \[ \text{Total momentum} = (m_1 + m_2) \cdot v \]
Substituting the known values into the equation: \[ 250 , \text{kg⋅m/s} = (1.5 , \text{kg} + m_2) \cdot 50 , \text{m/s} \]
Now, solve for \(m_2\):
First, divide both sides by \(50 , \text{m/s}\): \[ \frac{250 , \text{kg⋅m/s}}{50 , \text{m/s}} = 1.5 , \text{kg} + m_2 \] \[ 5 , \text{kg} = 1.5 , \text{kg} + m_2 \]
Next, isolate \(m_2\): \[ m_2 = 5 , \text{kg} - 1.5 , \text{kg} = 3.5 , \text{kg} \]
Therefore, the mass of the second object is:
3.5 kg.
To find the velocity of the ball after it is hit by the cue stick, we can use the conservation of momentum principle. In this scenario, we will assume that the cue stick transfers some of its momentum to the ball.
Before the collision:
Calculating initial momentum: The total initial momentum of the system is the momentum of the cue stick plus the momentum of the ball: \[ \text{Total initial momentum} = m_1 v_1 + m_2 v_2 \] \[ = (0.5 , \text{kg} \cdot 2.5 , \text{m/s}) + (0.2 , \text{kg} \cdot 0 , \text{m/s}) = 1.25 , \text{kg⋅m/s} \]
After the collision: Let \(v\) be the final velocity of the ball after being hit by the cue stick. The momentum of the system must be conserved, so we set the total initial momentum equal to the total final momentum.
The total final momentum is the momentum of the cue stick (which continues moving) and the momentum of the ball: \[ \text{Total final momentum} = m_1 v_{1f} + m_2 v \] Here, \(v_{1f}\) is the final velocity of the cue stick after the collision. We don't have this value, but we can set up the equation focusing on the ball's momentum.
Assuming that after the collision the cue stick continues moving with a reduced speed, we don't need the exact value of \(v_{1f}\) to answer this question. Solving for \(v\) requires either additional information or making assumptions, such as whether the collision is elastic or not. Without loss of generality, let's assume an elastic collision, so that momentum conservation can be effectively applied.
Assuming the cue stick slows down a bit but still has some speed \(v_{1f}\), we can apply an assumption for simplified analysis - that the cue stick stops transferring the bulk of its speed to the ball and the ball gains the majority of the impact momentum.
Using just the cue stick's initial speed, one could say approximately: \[ m_1 v_1 + m_2(0) = 0 + m_2 v \]
Setting for maximum conservation impact: \[ 1.25 = 0 + (0.2)v \]
Thus: \[ v = \frac{1.25}{0.2} = 6.25 , \text{m/s} \approx 6.3 , \text{m/s} \]
Therefore, the velocity of the ball after it is hit is approximately:
6.3 m/s.
The correct statement regarding the difference between assonance and consonance is:
Assonance relates to vowel sounds while consonance relates to consonant sounds.
Thus, they are both forms of sound device used in poetry and prose but focus on different types of sounds.
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