Question

A 50 kg cannonball hits a war tank vehicle of 12,000 kg that is initially at rest. The cannonball embeds in the tank and both move at the same velocity for 50 cm until they both come to a rest. The coefficient of friction between the tank and the pavement is 0.450. What was the initial velocity of the cannonball

Answer 1

Answer:

506.1 m/s

Explanation:

Let the initial velocity of cannonball be 'u' and final common velocity after collision be 'u₁'.

Given:

Mass of the cannonball (m) = 50 kg

Mass of the war tank (M) = 12,000 kg

Initial velocity of the tank (U) = 0 m/s

Distance moved by the system after collision (d) = 50 cm = 0.50 m

Coefficient of friction (μ) = 0.450

At the time of collision, momentum is conserved. So,

Initial momentum = Final momentum

$mu+MU=(m+M)u_1\\\\50u+0=(50+12000)u_1\\\\50u=12050u_1\\\\u_1=\frac{50}{12050}u\\\\u_1=\frac{u}{241}$

Now, as the system moves along the rough surface, the system eventually stops after traveling a certain distance.

The acceleration due to friction is given as:

$a=-\mu g\\\\a=-0.450\times 9.8=4.41\ m/s^2$

The negative sign implies that acceleration is acting to reduce the velocity.

Now applying the following equation of motion, we have:

$v_f^2 = v_i^2+2ad\\\\Where,v_f=0,v_i=u_1=\frac{u}{241}\ m/s, a=-4.41\ m/s^2,d=0.50\ m$

Now, solving for 'u', we get:

$0=(\frac{u}{241})^2-2\times 4.41\times 0.50\\\\\frac{u^2}{241^2}=4.41\\\\u^2=4.41\times 241^2\\\\u=\sqrt{4.41\times 241^2}\\\\u=506.1\ m/s$

Therefore, the initial velocity of the cannonball was 506.1 m/s.

Answer 2

The initial velocity of the cannonball would be:

506.1 m/s

Friction

According to the question,

Cannonball's mass, m = 50 kg

War tank's mass, M = 12,000 kg

Tank's initial velocity, U = 0 m/s

Distance after collision, d = 50 cm or,

                                          = 0.50 m

Friction coefficient, μ = 0.450

Let,

Cannonball's initial velocity be "u" and,

After collision, final common velocity be "u₁".

As we know,

→ Initial momentum = Final momentum

   mu + MU = (m + M)u₁

  50u + 0 = (50 + 12000)u₁

        50u = (50 + 12000)u₁

            u₁ = ($\frac{50}{12050}$)u

                = $\frac{u}{241}$

Acceleration will be:

→ a = -μg

     = - 0.450 × 9.8

     = 4.41 m/s²  

By applying equation of motion, we get

→ $v_f^2 =v_i^2 + 2ad$

    0 = ($\frac{u}{241}$)² - 2 × 4.41 × 0.50

   u² = 4.41 × (241)²

    u = √4.41 × 241²

       = 506.1 m/s

Thus the above answer is correct.

 

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https://brainly.com/question/24776317