Respuesta :
Answer:
17.34 m/s
Explanation:
Given:
Mass of lion (m₁) = 109 kg
Initial speed of lion (v₁) = 80.0 km/h (Northward direction)
Mass of gazelle (m₂) = 39.0 kg
Initial speed of gazelle (v₂) = 78.5 km/h (Eastward direction)
Final velocity of both lion and gazelle is, [tex]v_f=?[/tex]
First, let us convert the speeds from km/h to m/s using the conversion factor.
We know that, 1 km/h = 5/18 m/s
Therefore,
[tex]v_1= 80.0\ km/h=80\times \frac{5}{18}=22.22\ m/s\\\\v_2=78.5\ km/h=78.5\times \frac{5}{18}=21.81\ m/s[/tex]
Now, the concept of conservation of total momentum is used here as this is a case of perfectly inelastic collision. In inelastic collision, the masses move together with same velocity after collision.
Here, as the lion and gazelle are moving in directions at right angles to each other, the vector sum of their momentums will give the net initial momentum of the system.
So, initial momentum is given as:
[tex]P_i=\sqrt{P_1^2+P_2^2}\\\\Where,\\\\P_1\to initial\ momentum\ of\ lion\\P_2\to initial\ momentum\ of\ gazelle[/tex]
Now, we calculate P₁ and P₂.
[tex]P_1=m_1v_1=(109\ kg)(22.22\ m/s) = 2421.98\ Ns\\\\P_2=m_2v_2=(39\ kg)(21.81\ m/s) = 850.59\ Ns[/tex]
Therefore, the net initial momentum of the system is given as:
[tex]P_i=\sqrt{(2421.98)^2+(850.59)^2}=2567\ Ns[/tex]
The final momentum of the system is given as:
[tex]P_f=(m_1+m_2)(v_f)\\\\P_f=(109+39)v_f\\\\P_f=148v_f[/tex]
From the law of conservation of momentum, the final momentum is equal to the initial momentum. So,
[tex]P_f=P_i\\\\148v_f=2567\\\\v_f=\frac{2567}{148}=17.34\ m/s[/tex]
Therefore, the final speed of the lion-gazelle system is 17.34 m/s
The final speed v of the lion‑gazelle system just after the lion attacks is 79.60km/hr
To get the final speed of the lion, we will use the law of conservation of momentum expressed as:
[tex]m_1u_1 + m_2u_2=(m_1+m_1)v[/tex]
m1 and m2 are the masses of the lion and the gazelle
u1 and u2 are the velocities of the lion and the gazelle
v is their final velocity after the collision
Given the following parameters
m1 = 109kg
m2 = 39kg
u1 = =80.0km/hr
u2 = 78.5km/hr
Substitute the given parameters into the formula as shown:
[tex]109(80) + 39(78.5)=(109+39)v\\8720+3061.5=148v\\11,781.5=148v\\v = \frac{11,781.5}{148}\\v= 79.60km/hr[/tex]
Hence the final speed v of the lion‑gazelle system just after the lion attacks is 79.60km/hr
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