2. A 7.00-g bullet, when fired from a gun into a 1.00-kg block of wood held in a
ID: 1396653 • Letter: 2
Question
2. A 7.00-g bullet, when fired from a gun into a 1.00-kg block of wood held in a vise, penetrates the block to a depth of 8.00 cm. This block of wood is next placed on a frictionless horizontal surface, and a second 7.00-g bullet is fired from the gun into the block. To what depth will the bullet penetrate the block in this case? 3. The mass of the Earth is 5.97 x 24 kg, and the mass of the Moon is 7.35 x 10 22 kg. The distance of separation, measured between their centers, is 3.84 x10 m. Locate the center of mass of the Earth-Moon system as measured from the center of the Earth. 4. A small block of mass m1 -0.500 kg is released from W rest at the top of a frictionless, curve-shaped wedge of mass m2 E3.00 kg, which sits on a frictionless, horizontal surface as shown in Figure. When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right as shown in Figure b. (a) What is the velocity of the wedge after the block reaches the horizontal surface? (b) What is the height h of the wedge? mgo magExplanation / Answer
here,
mass of bullet , m = 0.007 kg
mass of block = 1 kg
when block is held in a vice,
final speed of bullet , V = 0
let the initial speed of bullet be u
depth , s = 0.08 m
by third equation of motion
a = (v^2 -u^2)/ 2 * s
a = - u^2 / 0.16 .....1
for the collison of block and bullet
let the final speed of buleet and block be vb
then using conservation of momentum
mass of bullet * speed of bullet = (mass of bullet + mass of box) * vb
0.007 * u = (0.007 + 1) * vb
vb = 6.99* 10^-3 * u
then after collison
initial speed of bullet = u
final speed of bullet , vb = 6.99* 10^-3 * u
accelration = - u^2 / 0.16
using newton third equation of motion
a = (v^2 -u^2)/ 2 * s
- u^2 / 0.16 = - ((6.99 * 10^-3)^2 * u^2 - u^2) /(2 * s)
s = 0.0799 m
the depth to which the bullet will penetrate this block is 0.0799 m
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