You are working on designing a carnival game where you throw 0.15kg bean bags at

Question

You are working on designing a carnival game where you throw 0.15kg bean bags at 0.35kg wooden targets and try to knock the wooden targets backwards across a goal line. You want to figure out where you can put the goal line so the block just barely reaches the line. Assume that when the bean bag hits the block it knocks the block over and lands on top of the block with both the bean bag and block moving together as the block slides along the gaming table. The coefficient of friction between the table and block is 0.3. You assume that a strong pitcher could throw a bean bag at roughly 24m/s. For simplicity, assume that the bean bag is traveling horizontal and hits the block head on. How fast are the bean bag and block moving immediately after they collide (but before they have slowed down due to sliding) assuming they stick together? How far do the block and bean bag slide after they collide?

Explanation / Answer

here ,

mass of bean bags , m = 0.15 kg

mass of wooden target , M = 0.35 kg

let the speed of target after collision is v m/s

Using conservation of momentum

(m + M ) * v = m * u

(0.15 + 0.35) * v = 0.15 * 24

v = 7.2 m/s

Now, for coefficient of friction , us = 0.3

acceleration due to friction, a = -0.3 * 9.8

a = -2.94 m/s^2

let the distance is d

Using third equation for motion

0 - 7.2^2 = 2 * (-2.94)* d

d = 8.82 m

the goal line should be 8.82 m from the targets for just touching the line

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