Asked by Sarah
A bullet of mass 12.4 g is fired into an initially stationary block and comes to rest in the block. The block, of mass 1.02 kg, is subject to no horizontal external forces during the collision with the bullet. After the collision, the block is observed to move at a speed of 4.80 m/s.
(a) Find the initial speed of the bullet.
(b) How much kinetic energy is lost?
(a) Find the initial speed of the bullet.
(b) How much kinetic energy is lost?
Answers
Answered by
Elena
The law of conservation of linear momentum for inelastic collision
m•v =(m +M) •u
v =(m +M) •u/m =(0.0124+1.02) •4.8/0.0124 = 399.6 m/s
KE1 =m•v^2/2 =0.0124•(399.6)^2/2 =990 J
KE2 = (m+M) •u^2/2=11.9 J
KE1 –KE2 =990 -11.9 = 978.1J
m•v =(m +M) •u
v =(m +M) •u/m =(0.0124+1.02) •4.8/0.0124 = 399.6 m/s
KE1 =m•v^2/2 =0.0124•(399.6)^2/2 =990 J
KE2 = (m+M) •u^2/2=11.9 J
KE1 –KE2 =990 -11.9 = 978.1J
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.