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A skateboarder is riding their skateboard at a speed of 0.3 m/s. They jump backwards off of the skateboard, resulting in the person coming to a stop horizontally, with the skateboard shooting off forwards. The skateboard has a mass of 4.9 kg, and the skateboarder has a mass of 42.9 kg. How fast would the skateboard be travelling after the skateboarder has jumped backwards off of it? U₁1

A skateboarder is riding their skateboard at a speed of 0.3 m/s. They jump backwards off of the skateboard, resulting in the person coming to a stop horizontally, with the skateboard shooting off forwards. The skateboard has a mass of 4.9 kg, and the skateboarder has a mass of 42.9 kg. How fast would the skateboard be travelling after the skateboarder has jumped backwards off of it? U₁1

University Physics Volume 1
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter9: Linear Momentum And Collisions
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A skateboarder is riding their skateboard at a speed of 0.3 m/s. They jump backwards off of the skateboard, resulting in the person coming to a stop horizontally, with the skateboard shooting off forwards. The skateboard has a mass of 4.9 kg, and the skateboarder has a mass of 42.9 kg. How fast would the skateboard be travelling after the skateboarder has jumped backwards off of it? 59 ··
Transcribed Image Text:A skateboarder is riding their skateboard at a speed of 0.3 m/s. They jump backwards off of the skateboard, resulting in the person coming to a stop horizontally, with the skateboard shooting off forwards. The skateboard has a mass of 4.9 kg, and the skateboarder has a mass of 42.9 kg. How fast would the skateboard be travelling after the skateboarder has jumped backwards off of it? 59 ··
Given data -Initial speed of skateboarder and skateboard is : U=0.3 m/s -Mass of skateboard is : m = 4,9 kg -Mass of skateboarder is; M = 42.9 kg. -Final speed of skateboarder is: V=0 The final speed of skateboard is calculated as ; (m+ M). u = mv + MV (4.9kg + 42.9 kq) (0.3 m/s) = (4.9 kg) v + (42.9 kg) (0) (4.4kg) V = 14.34 kg⋅m/s v = 2.9265 m/s V = 2.93 m/s 1. The final speed of the Skateboard is 2.93 m/s
Transcribed Image Text:Given data -Initial speed of skateboarder and skateboard is : U=0.3 m/s -Mass of skateboard is : m = 4,9 kg -Mass of skateboarder is; M = 42.9 kg. -Final speed of skateboarder is: V=0 The final speed of skateboard is calculated as ; (m+ M). u = mv + MV (4.9kg + 42.9 kq) (0.3 m/s) = (4.9 kg) v + (42.9 kg) (0) (4.4kg) V = 14.34 kg⋅m/s v = 2.9265 m/s V = 2.93 m/s 1. The final speed of the Skateboard is 2.93 m/s
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