# Solving the Bullet-Block Impulse Problem

• VSCCEGR
In summary, the problem involves a 20-g bullet fired into a 4-kg block, becoming embedded in it. The bullet and block then move up an incline for 1.2 seconds before coming to a stop. The task is to determine the magnitude of the bullet's initial velocity and the magnitude of the impulsive force exerted by the bullet on the block. The solution involves using the conservation of momentum along the incline plane and considering the collision of the bullet and block as well as the slide up the incline. After solving for the initial velocity and using the law of conservation of momentum, the magnitude of the bullet's initial velocity is found to be 707m/s. The impulsive force can be calculated using
VSCCEGR
Problem:

A 20-g bullet is fired into a 4-kg block and becomes embeded in it. knowing that the bullet and block then move up the incline for 1.2 seconds before coming to a stop Determine; (a) the Magnitude Of the bullets initial velocity, (b) the Magnitude of the Impulsive force by the bullet on the block.

Vb_1=? mk=4kg t=1.2s
mb=.02kg

I want to use the Princ. Impulse n' Momentum.
I have tried:
sum(mv_1)=sum(mv_2) but this does not help much

I have also thought of using some sort of constant Accel. Eq. Again to no avail.

Send me in the right direction.

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use the conservation of momentum along the incline plane...

here are some hints:
what is the magnitude of v_0 along the incline plane?
what is the initial velocity of the block? (conservation of momentum)
after you have the initial velocity, the rest should be easy

V_o along the incline is (V_o/2)
Initial Vel. Of block is 0
But how does this apply?

Impulse & Momentum:
sum(mv_1)+sum(Imp_1,2)=sum(mv_2)
Right, So:
sum(mv_1)=mass bullet *v_o
sum(Imp_1,2)=.5F-sin(15)4kg*9.81m/s
sum(mv_2)=(mass bullet+mass block)*v_2

F being the force exerted by the bullet on the block
v_o being velocity of the bullet
v_2 being velocity of bullet and block

F is what is asked for in (b) part
but i need v_o first

VSCCEGR said:
V_o along the incline is (V_o/2)
Initial Vel. Of block is 0
$V_0$ along the incline (if I read your diagram correctly) is $V_0 \cos{30}$.

I would treat the problem in two steps:

(1) The collision of bullet and block: The collision is governed by conservation of momentum parallel to the incline.
(2) The slide up the incline: Assuming no friction, the "bullet + block" system starts with some intial speed and has an acceleration down the plane due to gravity.

Do part 2 first.

Since you are not given the time it takes for the bullet to imbed itself into the block, I don't see how you can calculate the impulsive force.

Doc Al said:
Since you are not given the time it takes for the bullet to imbed itself into the block, I don't see how you can calculate the impulsive force.

Thats what I thought.

So:
V_0=?
V=0
a=9.81m/s verticly or 2.54m/s down incline
m=4.02kg
t=1.2s

cos(30)V_o=velocity of the bullet slong incline

Last edited:
VSCCEGR said:
So:
V_0=?
V=0
a=9.81m/s verticly or 2.54m/s down incline
m=4.02kg
t=1.2s
Use the acceleration and time to solve for the initial speed of the system (post collision) using kinematics. (That's what I called part 2)

As DocAl suggested:

1. Determine the initial post-collision speed of the sytem. You know t, and acceleration ($$gsin\theta$$), so find $$v_0$$. Then with the law of conservation of momentum determine the magnitude of the bullet's initial velocity.

2. Are you sure that the question doesn't ask for the Impulse J? Or is collision time given?

Found (a):
V=Vo+at
0=Vo+2.54*1.2
Vo=3.04m/s
mv_1=mv_2
.02[V_1/cos(30)]=(4+.02)3.04

V_1=707m/s

as for (b)
i think they mean:
Imp_1,2=F(delta)t
so I am guessing they want to find F(delta)t as a whole using
m_1*v_1+F(delta)t=m_2*v_2

Last edited:
by the way would this problem be done in the same way?

4oz ball @ 9ft/s strikes a 10oz plate on springs with no loss of energy. find Vel. of ball after hitting plate, force exerted by ball on plate.

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Yes, it's the same idea except that in this case you have energy conservation (elastic collision). The ball bounces off the plate. You have the same issue with the force. All you can do is find the total impulse because you cannot know the time or F(t) involved.

VSCCEGR said:
as for (b)
i think they mean:
Imp_1,2=F(delta)t
so I am guessing they want to find F(delta)t as a whole using
m_1*v_1+F(delta)t=m_2*v_2

$$J = \Delta p_{block} = p_{final}$$

In the final momentum do not add the bullet's mass.

Found (a):
V=Vo+at
0=Vo+2.54*1.2
Vo=3.04m/s
mv_1=mv_2
.02[V_1/cos(30)]=(4+.02)3.04
V_1=707m/s

Sorry ramollari, but you do using this method 707m/s is correct.

VSCCEGR said:
Sorry ramollari, but you do using this method 707m/s is correct.

I think ramollari's comment is directed toward the new problem you asked about, not the original one.

OlderDan said:
I think ramollari's comment is directed toward the new problem you asked about, not the original one.

No OlderDan, I meant the original problem. Since VSCCEGR was looking for the impulse that the bullet applies on the block, then by definition it is the change of momentum of the block.

$$J = \Delta p = p_{final} - 0 = p_{final}$$

Yes. of course you are right. Same thing applies to both the old problem and the new. I am just getting confused by the sequencing of comments with some of these these add on problems. Thanks for the clarification.

## 1. What is the bullet-block impulse problem?

The bullet-block impulse problem is a physics problem that involves determining the force and velocity of a bullet after it collides with a block. It is used to study the concept of impulse and momentum conservation.

## 2. How do you solve the bullet-block impulse problem?

To solve the bullet-block impulse problem, you need to use the principles of impulse and momentum conservation. This involves setting up equations to calculate the initial and final momentum of the bullet and the block, and then solving for the unknown variables.

## 3. What factors affect the outcome of the bullet-block impulse problem?

The outcome of the bullet-block impulse problem is affected by factors such as the mass and velocity of the bullet and the block, the angle of impact, and the material of the block (e.g. wood, steel, etc.). Also, the presence of external forces such as friction can also impact the outcome.

## 4. Are there any real-world applications for solving the bullet-block impulse problem?

Yes, the bullet-block impulse problem has real-world applications in fields such as forensics, ballistics, and engineering. It can also be used to study the effects of collisions in sports and other activities.

## 5. Can the bullet-block impulse problem be solved analytically?

Yes, the bullet-block impulse problem can be solved analytically using equations and mathematical principles. However, in some cases, numerical methods may be used to solve the problem due to its complexity.

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