# Simple exercise about conservation of momentum [closed]

A block of wood of mass $$M$$ is dropped, with no initial speed, from a height $$h$$ with respect to the ground. When it is at altitude $$\frac{h}{2}$$ it is hit by a bullet of mass m that travels horizontally with speed v. After the impact the bullet remains embedded in the wood. Determine the coordinate of the ground impact point of the block + projectile system. Consider the wood block and the bullet as material points. Perform the calculations for: $$M = 1 kg$$, $$h = 10 m$$, $$m = 10 g$$, $$v = 800 \frac{m}{s}$$

My problem:

From conservation of momentum along the x axis I can write that formula(1): $$(m+M)V_x=mv$$ so $$V_x=(\frac{m}{m+M})v$$

There isn't conservation of momentum along y axis so I cannot write the following formula(2): $$(m+M)V_y=MV_{\frac{h}{2}}$$ where $$V_{\frac{h}{2}}$$ is velocity of mass M calculated from height $$\frac{h}{2}$$. $$V_{\frac{h}{2}}$$ from conservation of energy is simple to calculate, so $$V_{\frac{h}{2}}=\sqrt{gh}$$

The solution of this exercise write formula $$(1)$$ and formula $$(2)$$. I'm not able to understand why I can write formula (2).

Thanks!

## closed as off-topic by G. Smith, stafusa, Gert, Aaron Stevens, Jon CusterAug 26 at 12:55

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• Why do you think there is no momentum conservation in the vertical? The question has given no indication of this. – Paul Childs Aug 26 at 3:14
• Try writing it as $lim_{\epsilon->0} (m+M)v_{h/2 + \epsilon}=mv_{h/2-\epsilon}$ – Paul Childs Aug 26 at 3:20
• @PaulChilds Because I know that there is momentum conservation when the sum of external forces is 0. On the vertical we have force of gravity. – ABC Aug 26 at 10:19

The change in the vertical momentum due to the force of gravity over a time interval $$\Delta t$$ is given by $$\Delta \vec p = \vec f_g\Delta t$$. If the impact of the bullet is instantaneous, you can write a conservation of momentum equation between before and after, assuming $$\Delta t=0$$ between before and after impact.

You actually do not even need to apply the conservation of linear momentum along y axis in this question. However if the bullet had some y component too, even then you could apply conservation of linear momentum along y-axis, because since both the bodies have the same acceleration (g), therefore they have no relative downward acceleration wrt each other. Hence, at the moment of collision, the conservation of linear momentum is valid even along the y axis.

Mass is a scalar quantity and velocity is a vector quantity. Due to this, momentum is also a vector quantity. This means that momentum can be resolved into components, and conservation of momentum applies to each component direction.

Given the fact that conservation of momentum applies to both the x and y directions, it is seen that you need to use conservation of momentum only in the x direction regarding the bullet, because that bullet is moving horizontally when it strikes the block. Since there is no vertical component of velocity for the bullet, the vertical velocity that is calculated based only on straight-line kinematics or conservation of energy for the vertical direction, is the correct physics description for velocity in that direction.

• So I didn't understand, does formula (2) make sense or not? – ABC Aug 25 at 23:22
• Since in my case, along the y-axis, there is the force of gravity, along this axis the momentum is not conserved right? – ABC Aug 25 at 23:25
• Formula 2 makes sense. There is no conservation of momentum in the vertical direction because there is no vertical component of momentum involved with the bullet collision. – David White Aug 26 at 2:26
• @Stochastic's answer makes sense. So, ABC you can apply conservation of linear momentum for y-axis, taking the vertical component of the instantaneous velocity, as collision is only momentary. – user233565 Aug 26 at 3:01
• You don't use conservation of energy. The fact that the bullet has embedded in the block means you have an unknown energy change due to deformation. Momentum conservation in the vertical direction is used as @stochastic has rightly pointed out. – Paul Childs Aug 26 at 3:13