1)A bullet of mass 5.3 g is fired horizontally into a 3.7 kg wooden block at res

Question

1)A bullet of mass 5.3 g is fired horizontally into a 3.7 kg wooden block at rest on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.28. The bullet stops in the block, which slides straight ahead for 2.6 m (without rotation). (a) What is the speed of the block immediately after the bullet stops relative to it? (b) At what speed is the bullet fired?

2)A 5.81 kg box sled is coasting across frictionless ice at a speed of 2.44 m/s when a 10.2 kg package is dropped into it from above. What is the new speed of the sled?

Explanation / Answer

1):  The bullet and block stops at a distance of 2.6m

The acceleration due to friction a is - µg = -0.28*9.8 = - 2.744 m/s²

a)

Using the kinematic equation v² - u² = 2as

0 - u² = -2*2.744*2.6

-u2=-14.2688

Therefore u2=14.2688

the speed of the block immediately after the bullet stops is

u = 3.777m/s

b)

By conservation of momentum

m u = (m +m) u


0.0053*u = (0.0053+3.7)(3.777)

u1=13.994/0.0053

=2640.5
the bullet fired speed is

u = 2640.5 m/a

2)

. We have,

5.81xv=10.2xV

Given v=2.44m/s

5.81x2.44=10.2xV

V=14.1764/10.2

=1.389m/s

the new speed of the sled is V=1.389m/s

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.