/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 A wooden block of mass \(0.9 \ma... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 32

A wooden block of mass \(0.9 \mathrm{~kg}\) is suspended from the ceiling of a room by a long thin wire. A bullet of mass \(0.1 \mathrm{~kg}\) moving horizontally with a speed of \(100 \mathrm{~m} / \mathrm{s}\) strikes the block and gets embedded in it. The height to which the block rises will be \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(2.5 \mathrm{~m}\) (B) \(5.0 \mathrm{~m}\) (C) \(7.5 \mathrm{~m}\) (D) \(10.0 \mathrm{~m}\)

Short Answer

Expert verified
The height to which the block rises is (B) \(5.0 \mathrm{~m}\).

Step by step solution

01

Analyze the collision of the bullet with the block

Since there are no external horizontal forces acting on the bullet and the block, linear momentum will be conserved in the horizontal direction during the collision. Let's denote the initial velocities of the block and bullet before the collision as \(v_{B0}\) and \(v_{b0}\) respectively, and their final velocity after the collision as \(v_f\). Since the block is initially at rest, \(v_{B0} = 0\).
02

Apply conservation of momentum to the bullet and block

Let's apply the conservation of linear momentum in the horizontal direction. The initial momentum of the block-bullet system is given by \((m_B+m_b)v_f = m_Bv_{B0} + m_bv_{b0}\). We know the mass of the block, \(m_B= 0.9~kg\), the mass of the bullet, \(m_b= 0.1~kg\), and the initial speed of the bullet, \(v_{b0} = 100~m/s\). We can now calculate the final velocity, \(v_f\).
03

Calculate the final velocity of the block-bullet system

Substituting the values into the conservation of momentum equation, we get \((0.9+0.1)v_f = 0 + (0.1)(100)\). Solving for \(v_f\), we obtain \(v_f = \frac{10}{1} = 10~m/s\). The block and the embedded bullet together have a velocity of \(10~m/s\) after the collision.
04

Analyze the block's motion after the collision

Now we will analyze the block's motion vertically upward after the collision. According to the conservation of mechanical energy, the initial kinetic energy of the block-bullet system will be converted into potential energy when the system reaches its highest point, and the vertical velocity is momentarily zero. At the highest point, the increase in gravitational potential energy will be equal to the initial kinetic energy of the system.
05

Apply conservation of mechanical energy to the block-bullet system

Let's denote the height the block rises as \(h\). Then the conservation of mechanical energy equation will be given as \(\frac{1}{2}(m_B+m_b)v_f^2 = (m_B+m_b)gh\). We know the combined mass of the block and bullet, \(m_B+m_b = 1~kg\), the final velocity after the collision, \(v_f=10~m/s\), and the acceleration due to gravity, \(g=10~m/s^2\). We can now calculate the height, \(h\).
06

Calculate the height the block rises

Substituting the known values into the conservation of mechanical energy equation, we get \(\frac{1}{2}(1)(10)^2 = (1)(10)h\). Solving for \(h\), we obtain \(h = \frac{50}{10} = 5~m\). Therefore, the height the block rises is \(5~m\). So, the correct answer is (B) \(5.0 \mathrm{~m}\).

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Most popular questions from this chapter

A bullet of mass \(m\) moving with velocity \(v\) strikes a suspended wooden block of mass \(M .\) If the block rises to a height \(h\), the initial velocity of the block will be (A) \(\sqrt{2 g h}\) (B) \(\frac{M+m}{m} \sqrt{2 g h}\) (C) \(\frac{m}{M+m} 2 g h\) (D) \(\frac{M+m}{M} \sqrt{2 g h}\)

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