Question

A cube of mass m slides without friction at speed vo = 10 m/s. It undergoes a perfectly elastic collision with the bottom tip ofa rod of length d and mass M 2m. The rod is pivoted about a frictionless axle through its center, and initially it hangsstraight , down and is at rest. What is the cube's velocity ( In m/s ) after collision.​

Answer 1

Answer:

Question 3.

Give the steps you will use to separate the variables and then solve the equation:

(a) 3n – 2 = 46

(b) 5m + 7 = 17

(c) 20p3=40

(d) 3p10=6

Solution:

(a) 3n – 2 = 46

⇒ 3n- 2 + 2 = 46+ 2 (adding 2 to both sides)

⇒ 3n = 48

⇒ 3n + 3 = 48 ÷ 3

Explanation:

please give brainliest answer

Answer 2

Final Answer:

The velocity (in m/s ) of the cube undergoing the elastic collision as shown in the attached figure is 2metre/second.

Given:

The cube of mass m is said to slide without friction at speed vo = 10 m/s.

It then undergoes a perfectly elastic collision with the bottom tip of the rod having length d and mass M = 2m.

The rod is stated to be pivoted about a frictionless axle through its center, and initially it is said to hang straight down and is at rest.

To Find:

The velocity (in m/s ) of the cube is to be calculated.

Explanation:

The vital concepts important for figuring out  the solution to this problem are as follows

• The law of conservation of energy indicates that the final energy is equal to the initial energy for any body or system.
• In a perfectly elastic collision, the momentum is conserved.
• The value of the variable in the quadratic equation $ax^2+bx+c=0$ is $x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

Step 1 of 4

As per the given statement in the problem and the attached figure, write the following parameters.

• The moment of inertia of the rod with mass 2m and height d is $I=\frac{1}{6}md^2$
• The momentum before collision is $L_i=mv_0\frac{d}{2}$
• The momentum after collision is $L_f=mv_f\frac{d}{2}+I\omega$
• The energy before collision is $E_i=\frac{1}{2} mv_0^2$
• The momentum after collision is $E_f=\frac{1}{2} mv_0^2+\frac{1}{2}I\omega^2$

Step 2 of 4

Using the information and the conservation laws, the following is derived

$\omega=3\frac{v_0-v_f}{d}$

Step 3 of 4

In continuation with the above calculations, derive the following.

$v_f^2-\frac{6}{5}v_0v_f+\frac{1}{5}v_0^2=0$

Using the related formula as described above, the value is

$v_f\\=\frac{3}{5}v_0\pm\frac{2}{5}v_0\\=v_0,\frac{1}{5}v_0$

Step 4 of 4

Since the initial velocity cannot be equal to the final velocity, the final velocity of the cube is

$v_f\\\\=\frac{1}{5}v_0\\\\=\frac{1}{5}\times 10\\\\=2\;m/sec$

Therefore, the required correct answer is the velocity 2m/sec.

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