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A 1.2 kg block of wood hangs motionless from strings. A 50 g bullet, traveling horizontally, strikes the block and becomes embedded inside the block. Immediately after the bullet becomes embedded in the block, the block is observed to have a speed of 8.0 m/s. What was the speed of the bullet before it hit the block?
A. 200 m/s.
B. 12 m/s.
C 98 m/s.
D 9,604 m/s.
E 57 m/s.

Respuesta :

abidemiokin

Answer:

A. 200m/s

Explanation:

Using the law of conservation of momentum expressed as;

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses of the object

u1 and u2 are the respective velocities

v is the common velocity

Given

m1 = 1.2kg

u1 = 0m/s (block is a stationary object)

m2 = 50g= 0.05kg

u2 = ?

v = 8.0m/s

Substitute the values into the formula and get u2 (speed of the bullet before hitting the block)

1.2(0)+0.05u2 = (1.2 + 0.05)(8)

0.05u2 = 1.25(8)

0.05u2 = 10

u2 = 10/0.05

u2 = 200m/s

Hence the speed of the bullet before it hit the block is 200m/s

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