A sand bag of mass 8.0\text{ kg} is suspended vertically by a light rope. A projectile of … — JAMB UTME

A sand bag of mass $8.0\text{ kg}$ is suspended vertically by a light rope. A projectile of mass $0.50\text{ kg}$ moving horizontally at a speed of $170\text{ m s}^{-1}$ strikes the sand bag and becomes embedded in it. What is the common speed, in $\text{m s}^{-1}$ , of the sand bag and the embedded projectile immediately after collision?

Answer: 10 m s^{-1}

Answer

The common speed of the sand bag and embedded projectile immediately after the collision is

10.0\text{ m s}^{-1}

According to the principle of conservation of linear momentum, total momentum before impact equals total momentum after impact. The initial momentum of the system is entirely from the projectile:

p_i = 0.50 \times 170 = 85\text{ kg m s}^{-1}

. After collision, both objects move together with a total mass of

0.50 + 8.0 = 8.5\text{ kg}

. Setting

8.5 v = 85

gives

v = 10\text{ m s}^{-1}

Step-by-Step Solution

State the conservation of linear momentum equation for an inelastic collision

m_1 u_1 + m_2 u_2 = (m_1 + m_2) v

Since no net external horizontal force acts on the system during impact, total linear momentum is conserved.

Substitute the given values into the momentum balance equation

(0.50\text{ kg})(170\text{ m s}^{-1}) + (8.0\text{ kg})(0\text{ m s}^{-1}) = (0.50\text{ kg} + 8.0\text{ kg}) v

The projectile embeds into the sand bag, so they move together with a single combined mass.

Solve the linear equation for the common final velocity

v

85 = 8.5 v \implies v = 10\text{ m s}^{-1}

Dividing total initial momentum by total combined mass yields the final speed.

Key Concept

Conservation of Linear Momentum in Completely Inelastic Collisions

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic