Chapter 07, Problem 30 GO One object is at rest, and another is moving. The two

Question

Chapter 07, Problem 30 GO One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 27 m/s. The masses of the two objects are 2.9 and 7.8 kg. Determine the final speed of the two-object system after the collision for the case (a) when the large-mass object is the one moving initially and the case (b) when the small-mass object is the one moving initially.

Chapter 07, Problem 30 GO One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 27 m/s. The masses of the two objects are 2.9 and 7.8 kg. Determine the final speed of the two-object system after the collision for the case (a) when the large-mass object is the one moving initially and the case (b) when the small-mass object is the one moving initially.

Chapter 07, Problem 30 GO One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 27 m/s. The masses of the two objects are 2.9 and 7.8 kg. Determine the final speed of the two-object system after the collision for the case (a) when the large-mass object is the one moving initially and the case (b) when the small-mass object is the one moving initially.

Explanation / Answer

mass m1=2.9kg

mass m2=7.8kg

velocity(v)=27m/s

A)the velocity of m2 is 27m/s

initial velocity of m1 is 0m/s

after inelastic collision they stick together and travel with common velocity

momentum is conserved

m2u2+m1u1=(m1+m2)v

v=(7.8x27)/(2.9+7.8)=19.7m/s

B)if u2=0 that is larger block is at rest and u1=27m/s

applying conservation of momentum

m1u1+m2u2=(m1+m2)v

v=2.9x27/(2.9+7.8)

=7.3m/s

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