Question

A 15kg object is shot 30 degrees above the horizontal at 20m/s. When the object is at the top of its flight path, it has a perfectly inelastic collision with a 5kg object going  10m/s in the negative y-direction. What is the total displacement in the x-direction from the objects initial position? (As much details as possible in the solution please! (: I know this is an energy, collision and projectile motion problem.)

5kg 10m/s 15kg 7 20m/s 15kg 20kg displacement?

Answer 1

5kg 10m/s 13 15kg 7 20m/s 15kg 20kg displacement?

Let M be the mass of object 1 = 15 kg

$\theta$ be the angle = 30⁰

^{u be the initial velocity of object 1 = 20 m/s}

^{h be the maximum height of object 1}

d be the net displacement

d₁ be the displacemet from A to B

d₂ be the displacement from B to C

d = d₁ + d₂

m be the mass of object 2 which collide with object 1 = 5 kg

(M+m) is the mass of object 3 (After collision of object 1 and object 2) = 20 kg

Let the horizontal velocity of object 1 before collision be u_Mx

and the vertical velocity of object 1 before collision be u_My

During Projectile motion at maximum height there will be only horizontal component of velocity

so

u_Mx = ucos($\theta$) = 17.321 m\s

u_My = 0

Let the horizontal velocity of object 1 before collision be u_mx = 0m/s

the vertical velocity of object 1 before collision be u_my = 10 m/s

Let the horizontal velocity of object 3 before collision be v_x

and the vertical velocity of object 1 before collision be v_y

By conservation of Momentum in horizontal direction at maximum height of object 1

$Mu_{Mx}+0=(M+m)v_{x}$

$15\times17.321=20v_{x}$

$v_{x}=13\;\;m/s$

By conservation of Momentum in vertical direction at maximum height of object 1

$mu_{my}+0=(M+m)v_{y}$

$5\times10=20v_{y}$

$v_{y}=2.5\;\;m/s$

Let v be the Net velocity of object 3 just after collision

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(13)^{2}+(2.5)^{2}}$

$v=\sqrt{169+6.25}=13.24\;\;m/s$

Maximum height of object 1

$h=\frac{u^{2}sin^{2}(\theta)}{2g}=\frac{400\times sin^{2}(30^{0})}{2\times 9.8}$

$h=5.1\;\;m$

The displacement from A to B

$d_{1}=\frac{v^{2}sin(2\theta)}{2g}=\frac{400\times sin(60^0)}{2\times9.8}$

$d_{1}=8.84\;\;m$

Let t be time taken by object 3 to reach from maximum height to point C

$h=v_{y}t+\frac{gt^{2}}{2}$

$5.1=2.5t+\frac{9.8t^{2}}{2}$

$5.1=2.5t+4.9t^{2}$

$4.9t^{2}+2.5t-5.1=0$

$t=0.8\;\;s$

$\boldsymbol{d_{2}=v_{x}t=13\times0.8=10.4\;\;m}$

Net displacement will be

$d=d_{1}+d_{2}=$$\boldsymbol{d=d_{1}+d_{2}=8.84+10.4=19.24\;\;m}$