Stopping a Bullet: Calculate umin and xf

[ThEmptyTree]

Homework Statement

A bullet of mass $m_1$ traveling horizontally with speed $u$ hits a block of mass $m_2$ that is originally at rest and becomes embedded in the block. After the collision, the block slides horizontally a distance $d$ on a surface with friction, and then falls off the surface at a height $h$ as shown. The coefficient of kinetic friction between the block and the surface is $\mu_k$. Assume the collision is nearly instantaneous and all distances are large compared to the size of the block. Neglect air resistance.

(a) What is $u_{min}$, the minimum speed of the bullet so that the block falls off the surface? Express your answer in terms of some or all of the following: $m_1, m_2, \mu_k, d, h$ and $g$ for the gravitational constant.

(b) Assume that the initial speed of the bullet $u$ is large enough for the block to fall off the surface. Calculate $x_f$ , the position where the block hits the ground measured from the bottom edge of the surface. Express your answer in terms of some or all of the following: $m_1, m_2, \mu_k, u, d, h$ and $g$.

Relevant Equations

Newton's 2nd Law : $$\overrightarrow{F}=m\overrightarrow{a}$$
Conservation of momentum for instantaneous collision: $$\overrightarrow{p_1}=\overrightarrow{p_2}$$

(a) $u_{min}=\big(1+\frac{m_2}{m_1}\big)\sqrt{2\mu_k g d}$

(b) $x_f=\sqrt{\frac{2h}{g}\Big(\big(\frac{m_1}{m_1+m_2}u\big)^2-2\mu_k g d\Big)}$

Can someone check please?

 

[SammyS]

Homework Statement:: A bullet of mass $m_1$ traveling horizontally with speed u hits a block of mass $m_2$ that is originally at rest and becomes embedded in the block. After the collision, the block slides horizontally a distance $d$ on a surface with friction, and then falls off the surface at a height $h$ as shown. The coefficient of kinetic friction between the block and the surface is $\mu_k$. Assume the collision is nearly instantaneous and all distances are large compared to the size of the block. Neglect air resistance.

(a) What is $u_{min}$, the minimum speed of the bullet so that the block falls off the surface? Express your answer in terms of some or all of the following: $m_1, m_2, \mu_k, d, h$ and $g$ for the gravitational constant.

(b) Assume that the initial speed of the bullet $u$ is large enough for the block to fall off the surface. Calculate $x_f$ , the position where the block hits the ground measured from the bottom edge of the surface. Express your answer in terms of some or all of the following: $m_1, m_2, \mu_k, u, d, h$ and $g$.
Relevant Equations:: Newton's 2nd Law : $$\overrightarrow{F}=m\overrightarrow{a}$$
Conservation of momentum for instantaneous collision: $$\overrightarrow{p_1}=\overrightarrow{p_2}$$

View attachment 288337

(a) $u_{min}=\big(1+\frac{m_2}{m_1}\big)\sqrt{2\mu_k g d}$

(b) $x_f=\sqrt{\frac{2h}{g}\Big(\big(\frac{m_1}{m_1+m_2}u\big)^2-2\mu_k g d\Big)}$

Can someone check please?

Explain how you arrived at those answers.

Please, show your work.

 

[haruspex]

(a) $u_{min}=\big(1+\frac{m_2}{m_1}\big)\sqrt{2\mu_k g d}$

(b) $x_f=\sqrt{\frac{2h}{g}\Big(\big(\frac{m_1}{m_1+m_2}u\big)^2-2\mu_k g d\Big)}$

Can someone check please?

Looks right.

 

[ThEmptyTree]

@haruspex Thanks for checking.

This is a sketch of what I've done:

(a)
$t=t_1:\text{right before the collision}$
$$\overrightarrow{p_1}=m_1\overrightarrow{u}$$
$t=t_2:\text{right after the collision}$
$$\overrightarrow{p_2}=(m_1+m_2)\overrightarrow{v_2}$$
Conservation of momentum to find $v_2$:
$$\overrightarrow{p_1}=\overrightarrow{p_2}\Rightarrow v_2=\frac{m_1}{m_1+m_2}u$$
Newton's 2nd law to find acceleration:
$$\overrightarrow{F}=m\overrightarrow{a}\Rightarrow a=-\mu_k g$$
Considering the case when the block stops on the edge:
$$v^2=v_0^2+2a(x-x_0)\Rightarrow 0=v_2^2+2ad\Rightarrow u_{min}=\big(1+\frac{m_2}{m_1}\big)\sqrt{2\mu_k g d}$$

(b)
Applying the same logic to find horizontal component of falling speed and so $x$ as a function of $t$:
$$v_x=\sqrt{\big(\frac{m_1}{m_1+m_2}u\big)^2-2\mu_k g d},~x=v_x t$$
Using kinematics to find $y$ as a function of $t$:
$$y=h-\frac{1}{2}gt^2$$
Eliminating $t$ from both equations:
$$y=h-\frac{g}{2v_x^2}x^2$$
At $y=0\Rightarrow x=x_f~:$
$$x_f=\sqrt{\frac{2h}{g}\Big(\big(\frac{m_1}{m_1+m_2}u\big)^2-2\mu_k g d\Big)}$$