Question

A bullet of mass m = 8.00 g is fired into a block of mass M = 230 g that is initially at rest at the edge of a table of height h = 1.00 m (see figure below). The bullet remains in the block, and after the impact the block lands d = 2.30 m from the bottom of the table. Determine the initial speed of the bullet.

________m/s

Answer 1

Block + bullet = 238g.
The block falls 1m., so from that you can work time in air.
Sqrt.([2x height]/ 9.8) = .452 secs.
In .452 secs, it travelled horizontally 2.3m.
(2.3/.452) = 5.088m/sec.
5.088 x 238g = 1211.06.
1211.06/8g = 151.38 m/sec. bullet velocity

Answer 2

help for this

An 8.00 g bullet is fired into a 210 g block that is initially at rest at the edge of a table of 1.00 m height?

Block + bullet = 218g.
The block falls 1m., so from that you can work time in air.
Sqrt.([2x height]/ 9.8) = .452 secs.
In .452 secs, it travelled horizontally 1.6m.
(1.6/.452) = 3.54m/sec.
3.54 x 218g = 771.68.
771.68/8g = 96.46m/sec. bullet velocity.

Answer 3

Block + bullet = 238g.
The block falls 1m., so from that you can work time in air.
Sqrt.([2x height]/ 9.8) = .452 secs.
In .452 secs, it travelled horizontally 2.30m.
(2.30/.452) = 5.08m/sec.
5.08 x 230g = 1168.4.

1168.4/8g = 146.05m/sec. bullet velocity.

Answer 4

Time taken for mass (m+M) to travel down a distance of h = sqrt (2*h/g) = sqrt ( 2* 1/9.8 ) = 0.45175 s
The distance travelled in horizontal direction in 0.45175 seconds is d = 2.3 m
So , the horizontal velocity = distance/time = 2.3/0.45175 = 5.091267 m/s
Velocity of (m+M) = 5.091267 m/s
So final momentum after collision = (m+M) * 5.091267 = (8 + 230) * 5.091267 = 1211.7215 gm.m/s
Initial momentum = m*V = 8*V
so by conservation of momentum, 8*V = 1211.7215
so V = Initial velocity of bullet = 151.465 m/s is the answer