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Chapter 4: Problem 518

A bullet of mass \(\mathrm{m}\) moving with velocity \(\mathrm{v}\) strikes a block of mass \(\mathrm{M}\) at rest and gets embedded into it. The kinetic energy of the composite block will be (A) \((1 / 2) \mathrm{mv}^{2} \times[\mathrm{M} /(\mathrm{m}+\mathrm{m})]\) (B) \((1 / 2) \mathrm{mv}^{2} \times[(\mathrm{m}+\mathrm{m}) / \mathrm{M}]\) (C) \((1 / 2) \mathrm{MV}^{2} \times[\mathrm{m} /(\mathrm{m}+\mathrm{M})]\) (D) \((1 / 2) \mathrm{mv}^{2} \times[\mathrm{m} /(\mathrm{m}+\mathrm{M})]\)

Short Answer

Expert verified
The short answer is: (D) \((1 / 2) \mathrm{mv}^{2} \times[\mathrm{m} /(\mathrm{m}+\mathrm{M})]\)

Step by step solution

01

1. Initial Momentum of the Bullet and Block

Initially, the bullet has a momentum of (m * v) while the block is at rest and has a momentum of zero. So, the initial total momentum (P_initial) is given by: P_initial = m * v
02

2. Final Momentum of the Composite Block

After the collision, the bullet and block move together as a single entity with mass (m + M). Let their final velocity be V. Since momentum is conserved during the collision, we can equate the initial and final momentum: P_initial = P_final m * v = (m + M) * V
03

3. Calculate the Final Velocity of the Composite Block

Now we need to solve the above equation for V: V = (m * v) / (m + M)
04

4. Find the Final Kinetic Energy of the Composite Block

Now that we have the final velocity V, we can find the final kinetic energy (K_final) of the composite block. The kinetic energy formula is: K_final = (1/2) * (Mass of Composite Block) * (Final Velocity)^2 K_final = (1/2) * (m + M) * V^2 Substitute the value of V that we found in step 3: K_final = (1/2) * (m + M) * [(m * v) / (m + M)]^2 After simplifying, we get: K_final = (1/2) * mv^2 * [m / (m + M)] Comparing our result to the given options, the correct answer is: (D) \((1 / 2) \mathrm{mv}^{2} \times[\mathrm{m} /(\mathrm{m}+\mathrm{M})]\)

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Most popular questions from this chapter

Natural length of a spring is \(60 \mathrm{~cm}\), and its spring constant is \(2000 \mathrm{~N} / \mathrm{m}\). A mass of \(20 \mathrm{~kg}\) is hung from it. The extension produced in the spring is..... $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)$ (A) \(4.9 \mathrm{~cm}\) (B) \(0.49 \mathrm{~cm}\) (C) \(9.8 \mathrm{~cm}\) (D) \(0.98 \mathrm{~cm}\)

A ball dropped from a height of \(4 \mathrm{~m}\) rebounds to a height of $2.4 \mathrm{~m}$ after hitting the ground. Then the percentage of energy lost is (A) 40 (B) 50 (C) 30 (D) 600

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