Question 1:

Momentum of the cannonball after firing
The momentum of the cannonball is its mass multiplied by its velocity.

Mass of cannonball m = 2 kg

Velocity of cannonball v = 100 m/s

p =m × v =2 kg×100 m/s=200 kg m/s

The correct answer is C. 200 kg m s^-1

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Question 2:

Total momentum of the system after firing
According to the law of conservation of momentum, the total momentum of a closed system remains constant. Since the initial momentum of the cannon-cannonball system was 0 (both were at rest), the total momentum after firing must also be 0.

The correct answer is A. 0 kg m s^-1

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Question 3:

Recoil velocity of the cannon
Based on the conservation of momentum:
Initial momentum = Final momentum
0=p_cannon + p_{cannonball
​}
0=(m_c×v_c)+(m_b × v_b)

We know the mass of the cannon (m_c) is 200 kg and the momentum of the cannonball (p_b) is 200 kg m/s. We need to find the velocity of the cannon (v_c):

0=(200 kg × v_c)+ 200 kg m/s
200 ×v_c = −200 kg m/s
v_c = − 200 / 200
v_c = −1 m/s

The negative sign indicates that the cannon moves in the opposite direction of the cannonball. The magnitude of the recoil velocity is 1 m/s.

The correct answer is D. 1 m/s

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Question 4:

Speed of A+B after the collision
This problem can be solved using the principle of conservation of momentum. The total momentum of the system before the collision is equal to the total momentum after the collision.

Initial Momentum:
p_initial =(m_A×v_A)+(m_B ×v_B)

p_initial =(4 kg×3 m/s)+(2 kg×0 m/s)
p_initial =12 kg m/s

Final Momentum:
Since the trolleys stick together, the final momentum is the combined mass multiplied by the new velocity (V).
p_final =(m_A×v_A)+(m_B ×v_B)

=(4 kg+2 kg)×v= 6v kg m/s

Conservation:
p_initial =p_final

12 kg m/s= 6v kg m/s
v= 12/6 =2 m/s

*These A.I. responses may be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
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Question 5:

Impulse of trolley A on Trolley B
Impulse is defined as the change in momentum of an object. The impulse of trolley A on trolley B is equal to the change in momentum of trolley B.

Initial momentum of B:
p_B,initial = m_B × v_B = 2 kg×0 m/s = 0 kg m/s

Final momentum of B:
p_B,final = m_B ×v_B = 2 kg×2 m/s = 4 kg m/s

Impulse:
Impulse=Δp =p_B,final - p_B,initial

Impulse=4 kg m/s−0 kg m/s=4 kg m/s

This corresponds to answer D.

*These A.I. responses may be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
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Question 6:

The meaning of an inelastic collision
In an inelastic collision, kinetic energy is not conserved. During the collision, some of the kinetic energy is converted into other forms of energy, such as heat, sound, or the energy used to deform the objects (in this case, the magnets sticking together). While the kinetic energy of the system changes, momentum is always conserved in a collision, whether it is elastic or inelastic.

B. kinetic energy is not conserved.

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Question 7:

Explanation
To solve this, we can use the relationship between momentum (p) and kinetic energy (k).

First, let's define the initial values for the first particle:

Mass: m

Momentum: p=mv

Kinetic Energy: k=½mv²

We can derive a formula that links kinetic energy and momentum:

From the momentum equation, we can express velocity (v) as:
v= p/m
Substitute this expression for v into the kinetic energy equation:
k=½m(p/m)²

k= p²/2m ( *this is in the IB data book)
​

So, the kinetic energy of the first particle is k= p²/2m

Now, let's analyze the second particle:

Mass: ½ m

Momentum: p (same as the first particle)

Kinetic Energy: k₂
​

Using the same derived formula for kinetic energy, we can find the kinetic energy of the second particle:
k_{2 =} p²/2m

k_{2 =} p²/2(½ m)

Comparing k₂ to k:

k₂ = p²/ m

k= p²/2m

​Therefore, k₂ =2× (p²/2m)
​
k₂=2k

The kinetic energy of the second particle is twice the kinetic energy of the first particle.

The correct answer is C. 2k.

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Question 8:

Impulse on the ball

Impulse is the change in momentum. Momentum is a vector, meaning its direction matters. Let's define the initial direction (to the right) as positive and the final direction (to the left) as negative.

Initial momentum (p_initial): The ball is moving to the right.
p_initial =m×v

Final momentum (p_final): The ball rebounds to the left with the same speed.
p_final =m×(−v)=−mv

Change in momentum (Δp):
Δp=p_final −p_initial

=(−mv)−(mv)=−2mv

The magnitude of the impulse is 2mv. The negative sign indicates the direction of the impulse is to the left, which makes sense as the wall pushes the ball in that direction.

The correct answer is D. 2mv.

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Question 9:

Force acting on the ball

Impulse is also equal to the force multiplied by the time over which the force acts. The formula is:
Impulse=F×t
From the previous question, we know the magnitude of the impulse is 2mv. Therefore,
F×t = 2mv
F = 2mv /t

The correct answer is D. 2mv/t.

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Question 10:

Collision type and force direction

• Collision Type: An elastic collision is one in which the total kinetic energy of the system is conserved. Since the ball rebounds with the same speed it had initially, its kinetic energy has not changed. This makes the collision perfectly elastic.

• Direction of Force: The wall is stopping the ball's rightward motion and then reversing its direction to the left. This requires a force acting to the left, which is the direction of the impulse calculated in question 8.

The correct answer is A. elastic, to the left.

*These A.I. responses may be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.