<p>A 1.6 kg block is attached to the end of a 2.0 m string to form a pendulum. T

Question

<p>A 1.6 kg block is attached to the end of a 2.0 m string to form a pendulum. The pendulum is released from rest when the string is horizontal. At the lowest point of its swing when it is moving horizontally, the block is hit by a 10 g bullet moving horizontally in the opposite direction. The bullet remains in the block and causes the block to come to rest at the low point of its swing. What was the magnitude of the bullet's velocity just before hitting the block?<br />Answer<br /><br /><br /><img class="mceBlock" src="https://s3.amazonaws.com/answer-board-image/49253814-81f7-49da-9824-055ceaa057d4.gif" border="0" alt="uploaded image" />1- 1.0 km/s<br />2- 1.4 km/s<br />3- 1.6 km/s<br />4- 1.2 km/s</p>

Explanation / Answer

Let's calculate the block's velocity at the bottom of its swing byconservation of energy. First of all, let's set the lowest point ofthe pendulum's swing to be the 0 of potential energy, i.e. h = 0.Then the pendulum is initially at a height of 2 m, so it has PE =mgh = 1.6 kg (g) ( 2.0 m)                                                                        = 3.2 kg 9.81 m/s^2                                                                         = 31.392 J.

When the pendulum reaches the lowest point its energy is allkinetic. Thus, we have

31.392 m/s^2 = 0.5mv^2
31.392 = 0.5 x1.6 kg v^2
v^2 = 31.392/0.8
v = sqrt(39.24)
v = 6.26 m/s

So the momentum of the block is mv = 1.6 kg (6.26 m/s )                                                          = 10.02 kg m/s

Thus, for the bullet to stop the block, it must have a momentum of10.02 kg m/s in the opposite direction, to make a total momentum of0. The bullet's mass is 0.01 kg, so we must solve

10.02 kg m/s = 0.01*v
                   v = 10.02 kg. m/s/0.01
                    v = 1002 m/s

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.