Question

a 5.5 g dart is fired into a block of wood with a mass of 22.6 g. the wood block is initially at rest on a 1.5 m tall post. after the collision, the wood block and dart land 2.5m from the base of the post. what is the initial speed of the dart?

Answer 1

From the problem alone we can say that the dart and the block of wood combined into a single object moving together at the end. With that clue we know that the collision is an inelastic collision. The formula of an inelastic collision is:

$m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}$

First let us sort out our given:Mass should be in kg to get the proper answer. Now let's assign m1 as the mass of the dart and m2 as the mass of the block. 
m1 = 5.5g

$5.5g$ x $\frac{1kg}{1000g}= 0.0055kg$

m2 = 22.6g

$22.6g$ x $\frac{1kg}{1000g}= 0.0226kg$

So now we settled that we can set our given as:
M1 = .0055 kg
v1i = ?
M2 = 0.0226 kg
v2i = 0 m/s
dx = 2.5 m
dy = -1.5 m

Now you can see that we have 2 unknowns: v1i and vf. We need the vf to solve for the initial velocity of the dart or object 1. We have other given to consider, so we can make use of that to get our missing vf. 

Now, vf is the horizontal velocity after the collision. We do this by first using the equations for projectiles considering that we have an x and y dimension to consider. We use the y dimension to get the x. 

dy = -1.5 m 

a = 9.8m/s^2

viy = 0 (take note that the initial vertical velocity is 0)

t = ?

We can use the UAM equations to solve for the time in the y-dimension (vertical) to get the horizontal velocity. 

$dy = v_{iy}t + \frac{1}{2} at^{2}$

$1.5 = (0)t+\frac{1}{2} (9.8)t^{2}$

$1.5 = \frac{1}{2} (9.8)t^{2}$

$\frac{(2)(1.5)}{9.8}=t^{2}$

$\frac{(3)}{9.8}=t^{2}$

$\sqrt{0.3061} = \sqrt{t^{2}$

$0.553s = t$

Now using this, we can get the horizontal (x-dimension) velocity using the formula:
$v_{x} =d_{x}t$ and our given earlier for the horizontal distance is 2.5m and we solved for time 0.553s. Let's put that into our equation:
$v_{x} =d_{x}t$
$v_{x} =(2.5m)(0.553s)$ 
$v_{x} =4.52m/s$

Now we finally have our vf or velocity after the collision. Now let's get back to the equation.

$m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}$

From this we can derive the equation for v1i by isolating it. 

$v_{1i}= \frac{((m_{1}+m_{2})v_{f})-(m_{2}v_{2i})}{m_{1}}$

Now let's put in all our given and what we solved:

$v_{1i}= \frac{((0.0055kg+0.0226kg)4.52m/s)-((0.0226kg)0m/s)}{0.0055kg}$

$v_{1i}= \frac{(0.0281kg)4.52m/s)}{0.0055kg}$

$v_{1i}= \frac{0.127012kg.m/s}{0.0055kg}$

$v_{1i}= 23.09m/s$

The initial speed of the dart is 23.09 m/s or 23.10 m/s.