1) A bullet of mass m = 8.00 g is fired into a block of mass M = 200 g that is i
ID: 1359341 • Letter: 1
Question
1) A bullet of mass m = 8.00 g is fired into a block of mass M = 200 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.
2) Gayle runs at a speed of 3.95 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After she has descended a vertical distance of 5.00 m, her brother, who is initially at rest, hops on her back and together they continue down the hill. What is their speed at the bottom of the hill if the total vertical drop is 15.0 m? Gayle's mass is 48.0 kg, the sled has a mass of 5.10 kg and her brother has a mass of 30.0 kg.
Explanation / Answer
(1) First of all, we try to understand complete process.
So bullet was moving with some initial velocity say Vi and then they get embedded into the block and then they both move with some velocity V.
Now the block has covered d = 2.3 m horizontal distance and 1 m vertical distance.
So let us calculate the time to cover 1m vertical distance
S = ut + (1/2)at2 (considering only vertical motion)
S = 1 m , u is initial velocity in vertical direction = 0 , a is acceleration = g and t is time
1 = 0.5gt2
t = 0.451 seconds
SO in this time block has also covered the horizontal distance d
In horizontal direction since there is no acceleration therefore velocity remain constant that is V
d= V*t
V = 2.3 /0.451 = 5.0997 m/s
This is the velocity of the block when bullet get embedded into it.
Now applying conservationof mass
Say bullet initial speed is Vi
mBVi = (mB + mblock)V
where mB is the mass of bullet
8Vi = (200+8)5.0997
Vi = 132.59 m/s
so this will be the initial speed of the bullet.
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