Inquiry into Physics
Inquiry into Physics
8th Edition
ISBN: 9781337515863
Author: Ostdiek
Publisher: Cengage
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Chapter 3, Problem 2C

A bullet with a mass of 0.01 kg is tired horizontally into a block of wood hanging on a string. The bullet sticks in the wood and causes it to swing upward to a height of 0.1 m. If the mass of the wood block is 2 kg, what was the initial speed of the bullet?

Expert Solution & Answer
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To determine

The initial speed of the bullet.

Answer to Problem 2C

The initial speed of the bullet is 281.4m/s.

Explanation of Solution

Given:

Given that a bullet with a mass of 0.01kg is fired horizontally into a block of wood hanging on a string. The bullet sticks in the wood and causes it to swing upward to a height of 0.1m .Also, the mass of the wood block is 2kg

Formula used:

According to the law of conservation of momentum, when the two objects collide in an isolated system, the total momentum before and after the collision is same.

Calculation:

Given that the bullet sticks in the wood and swing upward to a height of 0.1m

Thus, kinetic energy of bullet is converted into the potential energy

Now, kinetic energy of combined mass just after the impact is calculated as

KE=12(mBullet+mWood)vBullet2

We have mBullet=0.01kg and mWood=2kg

Substituting the value in above formula, we get

KE=12(0.01kg+2kg)vBullet2 ------ Eq(1)

Also, potential energy of wood and bullet together is calculated as

PE=(mBullet+mWood)gh

We have h=0.1m, mBullet=0.01kg, mWood=2kg and g=9.8m/s2

Substituting the value in above formula, we get

PE=(0.01kg+2kg)(9.8m/s2)(0.1m) ------- Eq(2)

Equating Eq(1) and Eq(2), we get

12(0.01kg+2kg)vBullet2=(0.01kg+2kg)(9.8m/s2)(0.1m)

12(2.01kg)vBullet2=(2.01kg)(9.8m/s2)(0.1m)

vBullet2=(2.01kg)(9.8m/s2)(0.1m)×2(2.01kg)

vBullet2=1.96m2/s2

vBullet=1.4m/s

The impact of bullet on wooden block is an inelastic collision, thus momentum is conserved.

Using the conservation of momentum, we have

mBulletvInitial+mWoodvWood=(mBullet+mWood)vFinal

We have vBullet=vFinal=1.4m/s, mBullet=0.01kg, mWood=2kg and vWood=0m/s

Substituting the values, we have

(0.01kg)vInitial+(2kg)(0m/s)=(0.01kg+2kg)(1.4m/s)

(0.01kg)vInitial=(2.01kg)(1.4m/s)

vInitial=(2.01kg)(1.4m/s)(0.01kg)

vInitial=281.4m/s.

Conclusion:

Hence, the initial speed of the bullet is 281.4m/s.

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Chapter 3 Solutions

Inquiry into Physics

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