Question

a bullet of mass 0.25 kg released from a gun with qn initial velocity of 40ms¹ after hitting a wall it comes to rest in 20.5 what is the regarding force acting on it?​

Answer 1

Answer:

the solution is in the picture below

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Answer 2

Answer:

The force is 0.8N.

Explanation:

Given:

m = 0.25Kg

T = 12.5s

$V_{f}=40m/s\\ V_{i} =0m/s$

Here,

The change in momentum is denoted by I.

The initial momentum is denoted by $M_{i}$.

The final momentum is denoted by $M_{f}$.

The time is denoted by T.

The initial speed is denoted by $V_{i}$.

The final speed is denoted by $V_{f}$.

The force is denoted by F.

The mass is denoted by m.

Now,

By the equation,

$F=\frac{I}{T} \\F= \frac{I_{f} -I_{i} }{T}\\ F=\frac{mV_{f}- mV_{i} }{T} \\F =\frac{0.25\times40-0.25\times 0}{12.5} \\F=0.8N$

So, the force is 0.8N.