Question

An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of 1.00 m height (Fig. P6.26). The bullet remains in the block, and after the impact the block lands 2.00 m from the bottom of the table. Determine the initial speed of the bullet.

Answer 1

So the block fell through a height of 1 m in time t, where
1.0=(1/2)gt²
or
t=√(2/9.81)
=0.452 s

During this time, the block has travelled horizontally 2m, or
horiz. velocity
=2/0.452
=4.43 m/s

This velocity, v, is the velocity of the block after impact, given by the equation of momentum before and after impact, namely
m1u1+m2u2=(m1+m2)v
or
v=(m1u1+m2u2)/(m1+m2)
where
m1=8g
m2=250g
u1= to be determined
u2=0 (block)
Solve for u1 (velocity of bullet)

Answer 2

do the end first
mass m (does not matter what m is) lands 2 meters from the table

how long did it fall?
h = 0 at end
0 = 1 - (1/2)(9.8) t^2
2 = 9.8 t^2
t = .45 seconds to fall to floor off table = time in air
distance = speed*time
2 = speed* .45
speed = 4.4 m/s horizontal
.008 v = .258 (4.4)
v = 141.9 m/s

Answer 3

48．Consider a frictionless track as shown in Figure. A block of mass m1 = 5.00 kg is released from A. It makes a head-on elastic collision at B with a block of mass m2 =10.0 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision.

Answer 4

An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of 1.00 m height (Fig. P6.26). The bullet remains in the block, and after the i

Answer 5

I want answer for the above question

Answer 6

You have to use kinematics then you can use the collision equation. The kinematics equation you use is vfysq=viySq+2a🔻y
Collison equation is my+mv=mv+mv