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Maru [420]
3 years ago
14

Help help help HELP AAAAA

Physics
1 answer:
charle [14.2K]3 years ago
5 0
The answer is B

As seen on the graph, the bus maintains a 9m/s speed for a majority of the trip to school.
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A bullet with a mass m b = 11.9 mb=11.9 g is fired into a block of wood at velocity v b = 261 m/s. vb=261 m/s. The block is atta
fredd [130]

Answer:

0.372 kg

Explanation:

The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:

mu=(M+m)v (1)

where

m=11.9 g = 11.9\cdot 10^{-3}kg is the mass of the bullet

u=261 m/s is the initial velocity of the bullet

M is the mass of the block

v is the velocity at which the bullet and the block travels after the collision

We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write

\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2 (2)

where

k = 205 N/m is the spring constant

x = 35.0 cm = 0.35 m is the compression of the spring

From eq(1) we get

v=\frac{mu}{M+m}

And substituting into eq(2), we can solve to find the mass of the block:

(M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg

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Answer:

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