Math Is Fun Forum

tony123

Member

Registered: 2007-08-03

Posts: 229

A bullet

A bullet with a mass of 20 grams is horizontally fired at a body with a mass of 100 grams, suspended by a string with a length of 6 meters, . The bullet comes to rest in the body. Find the vertical distance that the combined system of the body and the bullet moves after the collision

KerimF

Member

From: Aleppo-Syria

Registered: 2018-08-10

Posts: 164

Re: A bullet

It seems you forgot to add the initial speed of the bullet.
I guess this speed is also assumed constant (no friction).

Last edited by KerimF (2024-01-10 19:28:18)

tony123

Member

Registered: 2007-08-03

Posts: 229

Re: A bullet

It seems you forgot to add the initial speed of the bullet.
I guess this speed is also assumed constant (no friction).

Im sorry v=58.8m/s

Bob

Administrator

Registered: 2010-06-20

Posts: 10,180

Re: A bullet

I think you can do this by energy.

First use mv = (m +M)V to calculate the velocity of the combined mass.

So initial kinetic energy = 1/2 (m + M) V^2

If it rises to a height of h above the start position, then all this energy is converted to potential energy = (m + M)gh.

My only worry about this is I haven't used the length of the string.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

KerimF

Member

From: Aleppo-Syria

Registered: 2018-08-10

Posts: 164

Re: A bullet

Kinetic energy = K.E. = m*v^2/2
Potential gravitational energy = P.E. = (m+M)*g*h

Based on the energy conservation, we can write (since the exercise is about an ideal situation):
m*v^2/2 = (m+M)*g*h
therefore:
h = m*v^2/(m+M)/g/2