What is the speed of the block after the collision?

Click For Summary
SUMMARY

The discussion focuses on calculating the speed of a block immediately after a bullet embeds into it, using the principles of physics. The bullet has a mass of 5.26 g, and the block has a mass of 2.40 kg. After the collision, the block travels 1.70 m on a surface with a coefficient of kinetic friction of 0.30. The correct approach involves equating kinetic energy and work done by friction, leading to the conclusion that the speed of the block immediately after the collision is 2.43 m/s.

PREREQUISITES
  • Understanding of kinetic energy (KE = 0.5mv²)
  • Knowledge of frictional force (Ffr = ukmg)
  • Basic principles of conservation of momentum
  • Familiarity with the concept of work-energy principle
NEXT STEPS
  • Study the conservation of momentum in inelastic collisions
  • Learn about the work-energy theorem and its applications
  • Explore the effects of varying coefficients of friction on motion
  • Investigate energy transformations in mechanical systems
USEFUL FOR

Students in physics, educators teaching mechanics, and anyone interested in understanding the dynamics of collisions and friction in motion.

huynhtn2
Messages
29
Reaction score
0

Homework Statement


When a bullet strikes a block of wood originally at rest, the bullet becomes embedded into the wood block which travels 1.70 m after the collision. If the mass of the bullet is 5.26 g and the mass of the block is 2.40 kg, what is the speed of the block immediately after the collision? The coefficient of kinetic friction between the block and the surface is 0.30.



Homework Equations


ke=0.5mv^2
Ffr= ukmg


The Attempt at a Solution


Ke = Ffriction
0.5mv^2= uk mg
v= square root (2(9.81)(0.3))
v= 2.43 m/s

i don't know what to do with the 1.70 m distance
 
Physics news on Phys.org
huynhtn2 said:

Homework Statement


When a bullet strikes a block of wood originally at rest, the bullet becomes embedded into the wood block which travels 1.70 m after the collision. If the mass of the bullet is 5.26 g and the mass of the block is 2.40 kg, what is the speed of the block immediately after the collision? The coefficient of kinetic friction between the block and the surface is 0.30.

Homework Equations


ke=0.5mv^2
Ffr= ukmg

The Attempt at a Solution


Ke = Ffriction
0.5mv^2= uk mg
v= square root (2(9.81)(0.3))
v= 2.43 m/s

i don't know what to do with the 1.70 m distance

Your second formula in the attempt at a solution is comparing energy and force. The units do not match. You can only compare an energy term with another energy term, or a force with another force.

How can you get an energy value using the friction force? Thinking about this will give you the solution.

Cheers -- sylas
 
Last edited:

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
7K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K