Respuesta :
A) 46.5 kg m/s
B) 3.72 m/s
C) -634.3 J
Explanation:
A)
The momentum of an object is a vector quantity given by:
[tex]p=mv[/tex]
where
m is the mass of the object
v is its velocity
Momentum is a vector quantity, so if we want to calculate the total momentum of a system of two blocks, we have to add the individual momenta vectorially.
Here we have:
- Momentum of block 1:
[tex]p_1 = m_1 v_1[/tex]
where
[tex]m_1 = 1.50 kg[/tex] is the mass
[tex]v_1=31.0 m/s[/tex] is the velocity
- Momentum of block 2:
[tex]p_2 = m_2 v_2[/tex]
where
[tex]m_2=11.0 kg[/tex] is the mass of block 2
[tex]v_2=0 m/s[/tex] is its velocity
So, the total momentum at the beginning is
[tex]p_i = p_1+p_2=(1.50)(31.0)+0=46.5 kg m/s[/tex]
B)
We can solve this problem by using the law of conservation of momentum.
In fact, for an isolated system (=no external forces), the total momentum of the system must be conserved.
Therefore, here we can write that the final total momentum is equal to the initial momentum calculated in part a):
[tex]p_f = p_i[/tex]
And the final momentum can be written as:
[tex]p_f = (m_1 + m_2)v_f[/tex]
where
[tex]m_1=1.50 kg[/tex] is the mass of block 1
[tex]m_2=11.0 kg[/tex] is the mass of block 2
[tex]v_f[/tex] is the final velocity of the two blocks sticking together
Combining the two equations and solving for vf, we find:
[tex]v_f = \frac{p_i}{(m_1 + m_2)}=\frac{46.5}{1.50+11.0}=3.72 m/s[/tex]
C)
Here we want to calculate the change in kinetic energy of the system.
The kinetic energy of an object is given by
[tex]K=\frac{1}{2}mv^2[/tex]
where
m is the mass of the object
v is its speed
Before the collision, the total kinetic energy of the system is just the kinetic energy of block 1 (which is the only one in motion), so:
[tex]K_i = \frac{1}{2}m_1 v_1^2 = \frac{1}{2}(1.50)(31.0)^2=720.8 J[/tex]
While the total kinetic energy after the collision is:
[tex]K_f = \frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}(1.50+11.0)(3.72)^2=86.5 J[/tex]
Therefore, the change in kinetic energy is
[tex]\Delta K=K_f-K_i=86.5-720.8=-634.3 J[/tex]
So, the system has lost 634.3 J of energy.
The type of collision in which the two blocks stick together after the
collision is a perfectly inelastic collision.
The correct responses are;
- A: The total initial momentum is 46.5 kg·m/s
- B: The final velocity is 3.72 m/s
- C: The change in kinetic energy in the two block system is -633.76 kJ
Reasons:
A: The magnitude of the total initial momentum, [tex]P_i[/tex], is given as follows;
[tex]P_i[/tex] = m₁·v₁ + m₂·v₂
Where;
v₂ = The initial velocity of block 2 = 0 (block 2 is at rest)
Therefore;
[tex]P_i[/tex] = 1.50 × 31.0 + 11.0 × 0 = 46.5
The total initial momentum, [tex]P_i[/tex] = 46.5 kg·m/s
B: According to the principle of conservation of linear momentum, we have;
Total initial momentum, [tex]P_i[/tex] = Total final momentum, [tex]P_f[/tex]
The blocks stick together after the collision, therefore, the final velocity, [tex]v_f[/tex], is the same
Which gives;
Total final momentum, [tex]P_f[/tex] = m₁·[tex]v_f[/tex] + m₂·[tex]v_f[/tex] = [tex]v_f[/tex]·(m₁ + m₂)
Therefore, [tex]P_i[/tex] = [tex]P_f[/tex], gives;
m₁·v₁ + m₂·v₂ = [tex]v_f[/tex]·(m₁ + m₂)
46.5 = [tex]v_f[/tex]·(1.50 + 11.0)
[tex]v_f = \dfrac{46.5}{1.5 + 11} = 3.72[/tex]
The final velocity, [tex]v_f[/tex] = 3.72 m/s
C: The change in kinetic energy, ΔK = [tex]\mathbf{K_{final} - K_{initial}}[/tex] is given as follows;
The kinetic energy, K.E. = [tex]\dfrac{1}{2} \cdot m\cdot v^2[/tex]
[tex]K_{final} = \dfrac{1}{2} \cdot (m_1 + m_2) \times v_f^2[/tex]
[tex]K_{intial} = \mathbf{\dfrac{1}{2} \cdot m_1 \cdot v_1^2}[/tex]
Therefore;
[tex]K_{final} = \dfrac{1}{2} \cdot (1.5 + 11.0) \times 3.72^2 = 86.49[/tex]
[tex]K_{intial} = \dfrac{1}{2} \times 1.50 \times 31^2 = 720.25[/tex]
ΔK = [tex]K_{final} - K_{initial}[/tex] = 86.49 - 720.25 = -633.76
The change in kinetic energy, ΔK = -633.76 kJ
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