A 10 g bullet moves at a constant speed of 500 m/s and collides with a 1.5 kg wo
ID: 2282551 • Letter: A
Question
A 10 g bullet moves at a constant speed of 500 m/s and collides with a 1.5 kg wooden block initially at rest. The surface of the table is frictionless and 70 cm above the floor level. After the collision the bullet becomes embedded into the block. The bullet-block system slides off the top of the table and strikes the floor.
a. Find the momentum of the bullet before the collision.
b. Find the kinetic energy of the bullet before the collision.
c. Find the velocity of the bullet-block system after the collision.
d. Find the kinetic energy of the bullet-block after the collision.
e. Find the change in kinetic energy during the collision.
f. How much time it takes the bullet-block system to reach the floor?
g. Find the maximum horizontal distance between the table and the striking point on the floor.
Here are the answers:a. 5 kg m/s b. 1250 J c. 3.3 m/s d. 8.2 J e. -1241 J f. 0.38s g. 1.25 m
However, I do not understand how to solve it and would appreciate any help. Thanks
Explanation / Answer
a) Pi = m1*u1
= 0.01*500
= 5 kg.m/s
b)KEi = 0.5*m1*u1^2
= 0.5*0.01*500^2
= 1250 J
c) m1*u1 = (m1+m2)*v
v = m1*u1/(m1+m2)
= 0.01*500/(0.01+1.5)
= 3.31 m/s
d) KEf = 0.5*(m1+m2)*v^2
= 0.5*(0.01+1.5)*3.31^2
= 8.275 J
e) delta KE = KEf - KEi
= 8.275 - 1250
= -1241.725 J
f) t = sqrt(2*H/g)
= sqrt(2*0.7/9.8)
= 0.378 s
g) Range = t*v
= 0.378*3.31
= 1.25 m
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.