A confused 53.kg bowler on ice skates is coasting forwards with speed 20 m/s when he catches a 6.3.kg bowling ball moving 6.m/s in the opposite direction as he. What is the bowler's speed immediately after making the catch? 18.15_m/s 17.24 m/s С. D. 16.58 m/s 20.26 m/s F. А. В. Е. 19.67_m/s 19.01_m/s

College Physics
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Problem Statement:**

A confused 53 kg bowler on ice skates is coasting forwards with a speed of 20 m/s when he catches a 6.3 kg bowling ball moving at 6 m/s in the opposite direction as he. What is the bowler’s speed immediately after making the catch?

**Options:**

A. 18.15 m/s  
B. 17.24 m/s  
C. 19.67 m/s  
D. 16.58 m/s  
E. 20.26 m/s  
F. 19.01 m/s  

**Explanation:**

This problem involves the conservation of momentum. The total momentum before the catch equals the total momentum after the catch. The formula for momentum is:

\[ \text{Total Momentum} = \text{mass} \times \text{velocity} \]

Before the catch, the combined momentum of the bowler and the ball is:

\[ (53 \, \text{kg} \times 20 \, \text{m/s}) + (6.3 \, \text{kg} \times (-6) \, \text{m/s}) \]

After the catch, the mass of the bowler and bowling ball is:

\[ (53 + 6.3) \, \text{kg} \]

The equation for the final speed \( v \) after the catch using conservation of momentum is:

\[ (53 \times 20) + (6.3 \times -6) = (53 + 6.3) \times v \]

Solving this will give the bowler's speed after making the catch.
Transcribed Image Text:**Problem Statement:** A confused 53 kg bowler on ice skates is coasting forwards with a speed of 20 m/s when he catches a 6.3 kg bowling ball moving at 6 m/s in the opposite direction as he. What is the bowler’s speed immediately after making the catch? **Options:** A. 18.15 m/s B. 17.24 m/s C. 19.67 m/s D. 16.58 m/s E. 20.26 m/s F. 19.01 m/s **Explanation:** This problem involves the conservation of momentum. The total momentum before the catch equals the total momentum after the catch. The formula for momentum is: \[ \text{Total Momentum} = \text{mass} \times \text{velocity} \] Before the catch, the combined momentum of the bowler and the ball is: \[ (53 \, \text{kg} \times 20 \, \text{m/s}) + (6.3 \, \text{kg} \times (-6) \, \text{m/s}) \] After the catch, the mass of the bowler and bowling ball is: \[ (53 + 6.3) \, \text{kg} \] The equation for the final speed \( v \) after the catch using conservation of momentum is: \[ (53 \times 20) + (6.3 \times -6) = (53 + 6.3) \times v \] Solving this will give the bowler's speed after making the catch.
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