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Chapter 11: Q. 49 (page 289)

a. A bullet of mass m is fired into a block of mass M that is at rest. The block, with the bullet embedded, slides distance dacross a horizontal surface. The coefficient of kinetic friction is 渭k. Find an expression for the bullet鈥檚 speed vbullet

b. What is the speed of a 10g bullet that, when fired into a 10kg stationary woodblock, causes the block to slide 5.0 cm across a wood table?

Short Answer

Expert verified

a. The expression of bullet's speed,

vbullet=mbullet+mblockmbullet2渭kgd

b. The speed of the bullet, vbullet=443m/s

Step by step solution

01

Part (a). Step 1. Given information

Mass of bullet is 10gand block is 10kgis given along with the initial velocities.

02

Step 2. Analyze and apply

Using the law of conservation of momentum we have,

mbulletvbullet+mblockvblock=(mbullet+block)vfvbullet=(mbullet+block)vmbullet

For the velocity after collision, we use the equation of motion,

vs,y2=vf2+2advf=-2ad=2渭kgd

03

Step 3. Substitute and solve

On substitution of the values we have,vbullet=mbullet+mblockmbullet2渭kgd

04

Part (b). Step 1. Substituting the values of the known parameter

On substituting the values of the known parameter we get the value of the velocity,

vbullet=0.01kg+10kg0.01kg2路0.2路9.81m/s2路0.05m=443m/s

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