Question

please answer and show work. Thanks!

A pellet of mass 2.51 g from a pellet-gun is fired horozontally at a tin can of mass 185 g poised on a post at a height of 1.2 m above the fround. The initial speed of the pellet is 91 m/s. The pellet passes through the can after hitting it and emerges with a speed of 71 m/s. The can is knocked off the post hotizontally. BEFORE AFTER Upon hitting the ground, how far, d, is the can from the post? d = x m

Answer 1

First, find the horizontal velocity of the can after the collision

During collision momentum is conserved

Initial momentum = Final momentum

$m_{pellet}*u_{pellet}+m_{can}*u_{can}=m_{pellet}*v_{pellet}+m_{can}*v_{can}$

$0.00251kg*91m/s+0.185kg*0m/s=0.00251kg*71m/s+0.185kg*v_{can}$

$0.00251*91+0.185*0=0.00251*71+0.185*v_{can}$

$0.22841=0.17821+0.185*v_{can}$

$0.22841-0.17821=0.185*v_{can}$

$\frac{0.22841-0.17821}{0.185}=v_{can}$

${\color{Red} v_{can}=0.27135m/s}$

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Consider the vertical motion of the can

Use formula $s=ut+\frac{1}{2}at^{2}$

$h=u_{y}t+\frac{1}{2}gt^{2}$

$1.2m=0m/s*t+0.5*9.81m/s^{2}*t^{2}$

$1.2=0*t+0.5*9.81*t^{2}$

$1.2=0.5*9.81*t^{2}$

$\sqrt{\frac{1.2}{0.5*9.81}}=t$

${\color{Red} t=0.495s}$

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Consider the horizontal motion of the can

There is no acceleration in horizontal direction, so horizontal velocity remains constant.

Horizontal velocity = Horizontal distance/time

$v_{can}=\frac{d}{t}$

$d=v_{can}*t$

$d=0.27135m/s*0.495s$

ANSWER: ${\color{Red} d=0.13432m}$

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