The change in momentum of m 1 during collision along x and y directions.

Question

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Answer 1

Question

Chapter 7, Problem 63P

(a)

To determine

The change in momentum of $m1$ during collision along x and y directions.

(a)

Expert Solution

Answer to Problem 63P

The change in momentum of $m1$ during collision along x and y directions are $−m1vi$ and $0.751m1vi$ respectively.

Explanation of Solution

Using the conservation of linear momentum along x direction,

$Δp1,x=m2(v2,x,i−v2,x,f)=m2(v2,x,i−14vicosθ)$

Using the conservation of linear momentum along y direction,

$Δp1,y=m2(v2,y,i−v2,y,f)=m2(v2,y,i−14visinθ)$

Conclusion:

Substitute 0 m/s for $v2,x,i$ , $5m1$ for $m2$ , $−36.9°$ for $θ$ to get $Δp1,x$ .

$Δp1,x=(5m1)(0−14vicos(−36.9°))=−0.9997m1vi≈−m1vi$

Substitute 0 m/s for $v2,y,i$ , $5m1$ for $m2$ , $−36.9°$ for $θ$ to get $Δp1,y$ .

$Δp1,x=(5m1)(0−14visin(−36.9°))=0.751m1vi$

Therefore, change in momentum of $m1$ during collision along x and y directions are $−m1vi$ and $0.751m1vi$ respectively.

(b)

To determine

The change in momentum of $m2$ during collision along x and y directions.

(b)

Expert Solution

Answer to Problem 63P

The change in momentum of $m2$ during collision along x and y directions are $m1vi$ and $−0.751m1vi$ respectively.

Explanation of Solution

Using the conservation of linear momentum along x direction,

$Δp2,x=m1(v1,x,i−v1,x,f)=m1(vi−v1,x,f)$

Using the conservation of linear momentum along y direction,

$Δp2,y=m1(v1,y,i−v1,y,f)=m1(v1,y,i−v1)$

Conclusion:

Substitute 0 m/s for $v1,x,f$ to get $Δp2,x$ .

$Δp2,x=m1(vi−0)=m1vi$

Substitute 0 m/s for $v1,y,i$ and $0.751vi$ for $v1$ to get $Δp2,y$ .

$Δp2,y=m1(0−0.751vi)=−0.751m1vi$

Therefore, change in momentum of $m2$ during collision along x and y directions are $m1vi$ and $−0.751m1vi$ respectively. The momentum changes in (a) and (b) are equal in magnitude and opposite in direction.

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Answer 2

Question

Chapter 7, Problem 63P

(a)

To determine

The change in momentum of $m1$ during collision along x and y directions.

(a)

Expert Solution

Answer to Problem 63P

The change in momentum of $m1$ during collision along x and y directions are $−m1vi$ and $0.751m1vi$ respectively.

Explanation of Solution

Using the conservation of linear momentum along x direction,

$Δp1,x=m2(v2,x,i−v2,x,f)=m2(v2,x,i−14vicosθ)$

Using the conservation of linear momentum along y direction,

$Δp1,y=m2(v2,y,i−v2,y,f)=m2(v2,y,i−14visinθ)$

Conclusion:

Substitute 0 m/s for $v2,x,i$ , $5m1$ for $m2$ , $−36.9°$ for $θ$ to get $Δp1,x$ .

$Δp1,x=(5m1)(0−14vicos(−36.9°))=−0.9997m1vi≈−m1vi$

Substitute 0 m/s for $v2,y,i$ , $5m1$ for $m2$ , $−36.9°$ for $θ$ to get $Δp1,y$ .

$Δp1,x=(5m1)(0−14visin(−36.9°))=0.751m1vi$

Therefore, change in momentum of $m1$ during collision along x and y directions are $−m1vi$ and $0.751m1vi$ respectively.

(b)

To determine

The change in momentum of $m2$ during collision along x and y directions.

(b)

Expert Solution

Answer to Problem 63P

The change in momentum of $m2$ during collision along x and y directions are $m1vi$ and $−0.751m1vi$ respectively.

Explanation of Solution

Using the conservation of linear momentum along x direction,

$Δp2,x=m1(v1,x,i−v1,x,f)=m1(vi−v1,x,f)$

Using the conservation of linear momentum along y direction,

$Δp2,y=m1(v1,y,i−v1,y,f)=m1(v1,y,i−v1)$

Conclusion:

Substitute 0 m/s for $v1,x,f$ to get $Δp2,x$ .

$Δp2,x=m1(vi−0)=m1vi$

Substitute 0 m/s for $v1,y,i$ and $0.751vi$ for $v1$ to get $Δp2,y$ .

$Δp2,y=m1(0−0.751vi)=−0.751m1vi$

Therefore, change in momentum of $m2$ during collision along x and y directions are $m1vi$ and $−0.751m1vi$ respectively. The momentum changes in (a) and (b) are equal in magnitude and opposite in direction.

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Answer 3

Question

Chapter 7, Problem 63P

(a)

To determine

The change in momentum of $m1$ during collision along x and y directions.

(a)

Expert Solution

Answer to Problem 63P

The change in momentum of $m1$ during collision along x and y directions are $−m1vi$ and $0.751m1vi$ respectively.

Explanation of Solution

Using the conservation of linear momentum along x direction,

$Δp1,x=m2(v2,x,i−v2,x,f)=m2(v2,x,i−14vicosθ)$

Using the conservation of linear momentum along y direction,

$Δp1,y=m2(v2,y,i−v2,y,f)=m2(v2,y,i−14visinθ)$

Conclusion:

Substitute 0 m/s for $v2,x,i$ , $5m1$ for $m2$ , $−36.9°$ for $θ$ to get $Δp1,x$ .

$Δp1,x=(5m1)(0−14vicos(−36.9°))=−0.9997m1vi≈−m1vi$

Substitute 0 m/s for $v2,y,i$ , $5m1$ for $m2$ , $−36.9°$ for $θ$ to get $Δp1,y$ .

$Δp1,x=(5m1)(0−14visin(−36.9°))=0.751m1vi$

Therefore, change in momentum of $m1$ during collision along x and y directions are $−m1vi$ and $0.751m1vi$ respectively.

(b)

To determine

The change in momentum of $m2$ during collision along x and y directions.

(b)

Expert Solution

Answer to Problem 63P

The change in momentum of $m2$ during collision along x and y directions are $m1vi$ and $−0.751m1vi$ respectively.

Explanation of Solution

Using the conservation of linear momentum along x direction,

$Δp2,x=m1(v1,x,i−v1,x,f)=m1(vi−v1,x,f)$

Using the conservation of linear momentum along y direction,

$Δp2,y=m1(v1,y,i−v1,y,f)=m1(v1,y,i−v1)$

Conclusion:

Substitute 0 m/s for $v1,x,f$ to get $Δp2,x$ .

$Δp2,x=m1(vi−0)=m1vi$

Substitute 0 m/s for $v1,y,i$ and $0.751vi$ for $v1$ to get $Δp2,y$ .

$Δp2,y=m1(0−0.751vi)=−0.751m1vi$

Therefore, change in momentum of $m2$ during collision along x and y directions are $m1vi$ and $−0.751m1vi$ respectively. The momentum changes in (a) and (b) are equal in magnitude and opposite in direction.

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Answer 4

Question

Chapter 7, Problem 63P

(a)

To determine

The change in momentum of $m1$ during collision along x and y directions.

(a)

Expert Solution

Answer to Problem 63P

The change in momentum of $m1$ during collision along x and y directions are $−m1vi$ and $0.751m1vi$ respectively.

Explanation of Solution

Using the conservation of linear momentum along x direction,

$Δp1,x=m2(v2,x,i−v2,x,f)=m2(v2,x,i−14vicosθ)$

Using the conservation of linear momentum along y direction,

$Δp1,y=m2(v2,y,i−v2,y,f)=m2(v2,y,i−14visinθ)$

Conclusion:

Substitute 0 m/s for $v2,x,i$ , $5m1$ for $m2$ , $−36.9°$ for $θ$ to get $Δp1,x$ .

$Δp1,x=(5m1)(0−14vicos(−36.9°))=−0.9997m1vi≈−m1vi$

Substitute 0 m/s for $v2,y,i$ , $5m1$ for $m2$ , $−36.9°$ for $θ$ to get $Δp1,y$ .

$Δp1,x=(5m1)(0−14visin(−36.9°))=0.751m1vi$

Therefore, change in momentum of $m1$ during collision along x and y directions are $−m1vi$ and $0.751m1vi$ respectively.

(b)

To determine

The change in momentum of $m2$ during collision along x and y directions.

(b)

Expert Solution

Answer to Problem 63P

The change in momentum of $m2$ during collision along x and y directions are $m1vi$ and $−0.751m1vi$ respectively.

Explanation of Solution

Using the conservation of linear momentum along x direction,

$Δp2,x=m1(v1,x,i−v1,x,f)=m1(vi−v1,x,f)$

Using the conservation of linear momentum along y direction,

$Δp2,y=m1(v1,y,i−v1,y,f)=m1(v1,y,i−v1)$

Conclusion:

Substitute 0 m/s for $v1,x,f$ to get $Δp2,x$ .

$Δp2,x=m1(vi−0)=m1vi$

Substitute 0 m/s for $v1,y,i$ and $0.751vi$ for $v1$ to get $Δp2,y$ .

$Δp2,y=m1(0−0.751vi)=−0.751m1vi$

Therefore, change in momentum of $m2$ during collision along x and y directions are $m1vi$ and $−0.751m1vi$ respectively. The momentum changes in (a) and (b) are equal in magnitude and opposite in direction.

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Answer 5

Chapter 7, Problem 63P

(a)

To determine

The change in momentum of $m1$ during collision along x and y directions.

(a)

Expert Solution

Answer to Problem 63P

The change in momentum of $m1$ during collision along x and y directions are $−m1vi$ and $0.751m1vi$ respectively.

Explanation of Solution

Using the conservation of linear momentum along x direction,

$Δp1,x=m2(v2,x,i−v2,x,f)=m2(v2,x,i−14vicosθ)$

Using the conservation of linear momentum along y direction,

$Δp1,y=m2(v2,y,i−v2,y,f)=m2(v2,y,i−14visinθ)$

Conclusion:

Substitute 0 m/s for $v2,x,i$ , $5m1$ for $m2$ , $−36.9°$ for $θ$ to get $Δp1,x$ .

$Δp1,x=(5m1)(0−14vicos(−36.9°))=−0.9997m1vi≈−m1vi$

Substitute 0 m/s for $v2,y,i$ , $5m1$ for $m2$ , $−36.9°$ for $θ$ to get $Δp1,y$ .

$Δp1,x=(5m1)(0−14visin(−36.9°))=0.751m1vi$

Therefore, change in momentum of $m1$ during collision along x and y directions are $−m1vi$ and $0.751m1vi$ respectively.

(b)

To determine

The change in momentum of $m2$ during collision along x and y directions.

(b)

Expert Solution

Answer to Problem 63P

The change in momentum of $m2$ during collision along x and y directions are $m1vi$ and $−0.751m1vi$ respectively.

Explanation of Solution

Using the conservation of linear momentum along x direction,

$Δp2,x=m1(v1,x,i−v1,x,f)=m1(vi−v1,x,f)$

Using the conservation of linear momentum along y direction,

$Δp2,y=m1(v1,y,i−v1,y,f)=m1(v1,y,i−v1)$

Conclusion:

Substitute 0 m/s for $v1,x,f$ to get $Δp2,x$ .

$Δp2,x=m1(vi−0)=m1vi$

Substitute 0 m/s for $v1,y,i$ and $0.751vi$ for $v1$ to get $Δp2,y$ .

$Δp2,y=m1(0−0.751vi)=−0.751m1vi$

Therefore, change in momentum of $m2$ during collision along x and y directions are $m1vi$ and $−0.751m1vi$ respectively. The momentum changes in (a) and (b) are equal in magnitude and opposite in direction.

Answer 6

Chapter 7, Problem 63P

(a)

To determine

The change in momentum of $m1$ during collision along x and y directions.

(a)

Expert Solution

Answer to Problem 63P

The change in momentum of $m1$ during collision along x and y directions are $−m1vi$ and $0.751m1vi$ respectively.

Explanation of Solution

Using the conservation of linear momentum along x direction,

$Δp1,x=m2(v2,x,i−v2,x,f)=m2(v2,x,i−14vicosθ)$

Using the conservation of linear momentum along y direction,

$Δp1,y=m2(v2,y,i−v2,y,f)=m2(v2,y,i−14visinθ)$

Conclusion:

Substitute 0 m/s for $v2,x,i$ , $5m1$ for $m2$ , $−36.9°$ for $θ$ to get $Δp1,x$ .

$Δp1,x=(5m1)(0−14vicos(−36.9°))=−0.9997m1vi≈−m1vi$

Substitute 0 m/s for $v2,y,i$ , $5m1$ for $m2$ , $−36.9°$ for $θ$ to get $Δp1,y$ .

$Δp1,x=(5m1)(0−14visin(−36.9°))=0.751m1vi$

Therefore, change in momentum of $m1$ during collision along x and y directions are $−m1vi$ and $0.751m1vi$ respectively.

Answer 7

Chapter 7, Problem 63P

(a)

To determine

The change in momentum of $m1$ during collision along x and y directions.

Answer 8

(a)

Expert Solution

Answer to Problem 63P

The change in momentum of $m1$ during collision along x and y directions are $−m1vi$ and $0.751m1vi$ respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Using the conservation of linear momentum along x direction,

$Δp1,x=m2(v2,x,i−v2,x,f)=m2(v2,x,i−14vicosθ)$

Using the conservation of linear momentum along y direction,

$Δp1,y=m2(v2,y,i−v2,y,f)=m2(v2,y,i−14visinθ)$

Conclusion:

Substitute 0 m/s for $v2,x,i$ , $5m1$ for $m2$ , $−36.9°$ for $θ$ to get $Δp1,x$ .

$Δp1,x=(5m1)(0−14vicos(−36.9°))=−0.9997m1vi≈−m1vi$

Substitute 0 m/s for $v2,y,i$ , $5m1$ for $m2$ , $−36.9°$ for $θ$ to get $Δp1,y$ .

$Δp1,x=(5m1)(0−14visin(−36.9°))=0.751m1vi$

Therefore, change in momentum of $m1$ during collision along x and y directions are $−m1vi$ and $0.751m1vi$ respectively.

Answer 13

(b)

To determine

The change in momentum of $m2$ during collision along x and y directions.

(b)

Expert Solution

Answer to Problem 63P

The change in momentum of $m2$ during collision along x and y directions are $m1vi$ and $−0.751m1vi$ respectively.

Explanation of Solution

Using the conservation of linear momentum along x direction,

$Δp2,x=m1(v1,x,i−v1,x,f)=m1(vi−v1,x,f)$

Using the conservation of linear momentum along y direction,

$Δp2,y=m1(v1,y,i−v1,y,f)=m1(v1,y,i−v1)$

Conclusion:

Substitute 0 m/s for $v1,x,f$ to get $Δp2,x$ .

$Δp2,x=m1(vi−0)=m1vi$

Substitute 0 m/s for $v1,y,i$ and $0.751vi$ for $v1$ to get $Δp2,y$ .

$Δp2,y=m1(0−0.751vi)=−0.751m1vi$

Therefore, change in momentum of $m2$ during collision along x and y directions are $m1vi$ and $−0.751m1vi$ respectively. The momentum changes in (a) and (b) are equal in magnitude and opposite in direction.

Answer 14

(b)

To determine

The change in momentum of $m2$ during collision along x and y directions.

Answer 15

(b)

Expert Solution

Answer to Problem 63P

The change in momentum of $m2$ during collision along x and y directions are $m1vi$ and $−0.751m1vi$ respectively.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Using the conservation of linear momentum along x direction,

$Δp2,x=m1(v1,x,i−v1,x,f)=m1(vi−v1,x,f)$

Using the conservation of linear momentum along y direction,

$Δp2,y=m1(v1,y,i−v1,y,f)=m1(v1,y,i−v1)$

Conclusion:

Substitute 0 m/s for $v1,x,f$ to get $Δp2,x$ .

$Δp2,x=m1(vi−0)=m1vi$

Substitute 0 m/s for $v1,y,i$ and $0.751vi$ for $v1$ to get $Δp2,y$ .

$Δp2,y=m1(0−0.751vi)=−0.751m1vi$

Therefore, change in momentum of $m2$ during collision along x and y directions are $m1vi$ and $−0.751m1vi$ respectively. The momentum changes in (a) and (b) are equal in magnitude and opposite in direction.